Two Dimensional Geometry jee mains questions | JEE Main Two Dimensional Geometry Previous Year Question

Q 1 :

A square is inscribed in the circle $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$ . One side of this square is parallel to y = x + 3. If $(x_{i}, y_{i})$ are the vertices of the square, then $\sum_{}^{} (x_{i}^{2}, y_{i}^{2})$ is equal to [2024]

160
152
156
148

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(2)

One side of square is y = x + k

Distance of (5, 3) to the line y = x + k is

$\frac{| 3 - 5 - k |}{\sqrt{2}} = \sqrt{2}$

$\Rightarrow | - 2 - k | = 2 \Rightarrow k = 0 or k = - 4$

So lines are y = x and y = x – 4

Now, solving these lines with circle

y = x and $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$

$\Rightarrow 2 x^{2} - 16 x + 30 = 0$

$\Rightarrow$ x = 3, y = 3 and x = 5, y = 5

and y = x – 4 and $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$

$\Rightarrow$ x = 5, y = 1 and x = 7, y = 3

$\sum_{i = 1}^{4} x_{i}^{2} + y_{i}^{2}$ = 9 + 9 + 25 + 25 + 25 + 1 + 49 + 9 = 152.

Q 2 :

Let C be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line x + y = 2 intersects the circle C at the points P and Q. Let MN be a chord of C of length 2 units and slope –1. Then a distance (in units) between the chord PQ and the chord MN is [2024]

$3 - \sqrt{2}$
$2 - \sqrt{3}$
$\sqrt{2} + 1$
$\sqrt{2} - 1$

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(1)

Distance of centre from chord $P Q =$ $| \frac{0 + 0 - 2}{\sqrt{1^{2} + 1^{2}}} |$ $= \frac{2}{\sqrt{2}} = \sqrt{2}$

Let distance of centre from chord MN = p, then length of chord = $2 \sqrt{10 - p^{2}}$

$\Rightarrow 2 \sqrt{10 - p^{2}} = 2$ ( $∵$ Given)

$\Rightarrow 10 - p^{2} = 1 \Rightarrow p^{2} = 9 \Rightarrow p = 3$

$\Rightarrow$ Distance between chord PQ and MN = $3 - \sqrt{2}$ .

Q 3 :

Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3, 2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5, 5) is: [2024]

$2 \sqrt{2}$
$4 \sqrt{2}$
4
5

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(3)

Centre of circle = (2, 1)

$∴$ Equation of circle will be

${(x - 2)}^{2} + {(y - 1)}^{2} = 1$

Distance between C(2, 1) and P(5, 5)

= $\sqrt{{(2 - 5)}^{2} + {(1 - 5)}^{2}}$

= 5 units

Also, AC = BC = 1 unit [ $∵$ Radius = 1]

$∴$ Shortest distance of circle from P = 5 – 1 = 4 units.

Q 4 :

Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at (–1, 0) and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the y-axis at the point P, then the square of the distance of P from the centre of $C_{1}$ is: [2024]

2
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(1)

We have, $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$

$C_{2} : {(x + 1)}^{2} + y^{2} - 4 = 0$

For common chord, we have $C_{1} - C_{2} = 0$

$\Rightarrow x^{2} + y^{2} - 2 x - 2 y + 1 - x^{2} - y^{2} - 2 x + 3 = 0$

$\Rightarrow - 4 x - 2 y + 4 = 0$

$\Rightarrow 2 x + y - 2 = 0$

Since, common chord intersects y-axis

So, x = 0

$\Rightarrow$ y = 2

So, point of intersection of common chord with y-axis is P(0, 2).

Required distance = ${(\sqrt{{(1 - 0)}^{2} + {(1 - 2)}^{2}})}^{2} = 2$ .

Q 5 :

Let ABCD and AEFG be squares of sides 4 and 2 units respectively. the point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies : [2024]

$2 r^{2} - 4 r + 1 = 0$
$r^{2} - 8 r + 8 = 0$
r = 1
$2 r^{2} - 8 r + 7 = 0$

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(2)

Let r be the radius of circle and centre at C(r, r).

Then, $F C^{2} = r^{2}$

$\Rightarrow {(r - 2)}^{2} + {(r - 2)}^{2} = r^{2}$

$\Rightarrow r^{2} - 8 r + 8 = 0$ .

Q 6 :

Let the circles $C_{1} : {(x - α)}^{2} + {(y - β)}^{2} = r_{1}^{2}$ and $C_{2} : {(x - 8)}^{2} + {(y - \frac{15}{2})}^{2} = r_{2}^{2}$ touch each other externally at the point (6, 6). If the point (6, 6) divides the line segment joining the centres of the circles $C_{1}$ and $C_{2}$ internally in the ratio 2 : 1, then $(α + β) + 4 (r_{1}^{2} + r_{2}^{2})$ equals [2024]

110
125
145
130

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(4)

$B (α, β), C (8, \frac{15}{2})$ are the centres of circle $C_{1}$ and $C_{2}$ respectively. Now, A(6, 6) is the given point.

A divides BC in ratio 2 : 1

$\Rightarrow \frac{16 + α}{3} = 6$ and $\frac{15 + β}{3} = 6$

$\Rightarrow α = 2 and β = 3$

Also, $\bar{B C} = r_{1} + r_{2} \Rightarrow r_{1} + r_{2} = \sqrt{{(2 - 8)}^{2} + {(3 - \frac{15}{2})}^{2}}$

$\Rightarrow 2 r_{2} + r_{2} = \sqrt{36 + \frac{81}{4}}$ $[∵ r_{1} : r_{2} = 2 : 1]$

$\Rightarrow 3 r_{2} = \frac{15}{2} \Rightarrow r_{2} = \frac{5}{2} units \Rightarrow r_{1} = 5 units$

$∴ (α + β) + 4 (r_{1}^{2} + r_{2}^{2}) = 5 + 4 (25 + \frac{25}{4}) = 130$ .

Q 7 :

If the image of the point (–4,5) in the line x + 2y = 2 lies on the circle ${(x + 4)}^{2} + {(y - 3)}^{2} = r^{2}$ , then r is equal to : [2024]

1
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(4)

$\frac{x + 4}{1} = \frac{y - 5}{2} = \frac{- 2 (- 4 + 2 \times 5 - 2)}{1 + 4} = \frac{- 8}{5}$

$\Rightarrow x = - 4 - \frac{8}{5} = - \frac{28}{5}, y = 5 - \frac{16}{5} = \frac{9}{5}$

$∴ Image is (\frac{- 28}{5}, \frac{9}{5})$

Image lies on circle ${(x + 4)}^{2} + {(y - 3)}^{2} = r^{2}$

${(\frac{- 28}{5} + 4)}^{2} + {(\frac{9}{5} - 3)}^{2} = r^{2}$

$\Rightarrow \frac{64}{25} + \frac{36}{25} = r^{2} \Rightarrow r^{2} = 4 \Rightarrow r = 2$

Q 8 :

Let a circle passing through (2,0) have its centre at the point (h, k). Let $(x_{c}, y_{c})$ be the point of intersection of the lines 3x + 5y = 1 and $(2 + c) x + 5 c^{2} y = 1$ . If $h = \lim_{c \to 1} x_{c}$ and $k = \lim_{c \to 1} y_{c}$ , then the equation of the circle is : [2024]

$5 x^{2} + 5 y^{2} - 4 x + 2 y - 12 = 0$
$5 x^{2} + 5 y^{2} - 4 x - 2 y - 12 = 0$
$25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0$
$25 x^{2} + 25 y^{2} - 2 x + 2 y - 60 = 0$

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(3)

We have, 3x + 5y = 1 ... (i)

$(2 + c) x + 5 c^{2} y = 1$ ... (ii)

Multiplying (i) by $c^{2}$ and subtracting it from (ii), we get

$(2 + c - 3 c^{2}) x = 1 - c^{2}$

$\Rightarrow x_{c} = \frac{1 - c^{2}}{2 + c - 3 c^{2}} = \frac{(1 - c) (1 + c)}{(1 - c) (3 c + 2)} = \frac{c + 1}{3 c + 2}$

$∴ y = \frac{1 - 3 x}{5} = \frac{1 - 3 (\frac{c + 1}{3 c + 2})}{5} = \frac{3 c + 2 - 3 c - 3}{5 (3 c + 2)}$

$y_{c} = \frac{- 1}{5 (3 c + 2)}$

Now, $h = \lim_{c \to 1} x_{c} = \lim_{c \to 1} (\frac{c + 1}{3 c + 2}) = \frac{1 + 1}{3 + 2} = \frac{2}{5}$

and $k = \lim_{c \to 1} y_{c} = \lim_{c \to 1} \frac{- 1}{5 (3 c + 2)} = \frac{- 1}{25}$

Equation of circle is

${(x - \frac{2}{5})}^{2} + {(y + \frac{1}{25})}^{2} = \frac{64}{25} + \frac{1}{625}$ [ $∵$ Circle passes through (2, 0)]

$\Rightarrow x^{2} + y^{2} - \frac{4}{5} x + \frac{2}{25} y + \frac{4}{25} + \frac{1}{625} = \frac{64}{25} + \frac{1}{625}$

$\Rightarrow 25 x^{2} + 25 y^{2} - 20 x + 2 y + 4 = 64$

$\Rightarrow 25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0$

Q 9 :

Let $C : x^{2} + y^{2} = 4$ and $C' : x^{2} + y^{2} - 4 λ x + 9 = 0$ be two circles. If the set of all values of $λ$ so that the circles C and C' intersect at two distinct points, is R – [a, b], then the point (8a + 12, 16b – 20) lies on the curve : [2024]

$x^{2} - 4 y^{2} = 7$
$6 x^{2} + y^{2} = 42$
$5 x^{2} - y = - 11$
$x^{2} + 2 y^{2} - 5 x + 6 y = 3$

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(2)

We have, $C : x^{2} + y^{2} = 4$ ... (i)

and $C' : x^{2} + y^{2} - 4 λ x + 9 = 0$

$\Rightarrow C' : {(x - 2 λ)}^{2} + y^{2} = 4 λ^{2} - 9$ ... (ii)

Radius of C, $r_{1} = 2$ and radius of C', $r_{2} = \sqrt{4 λ^{2} - 9}$

When two circles intersect at two points,

then $| r_{1} - r_{2} | < C C' < | r_{1} + r_{2} |$

$\Rightarrow$ $| 2 - \sqrt{4 λ^{2} - 9} |$ $< 2 λ < 2 + \sqrt{4 λ^{2} - 9}$ ...(iii)

By (iii), we have $4 + 4 λ^{2} - 9 - 4 \sqrt{4 λ^{2} - 9} < 4 λ^{2}$

$\Rightarrow - 5 - 4 \sqrt{4 λ^{2} - 9} < 0$

$\Rightarrow 5 + 4 \sqrt{4 λ^{2} - 9} > 0 \Rightarrow \sqrt{4 λ^{2} - 9} > - \frac{5}{4}$

On Squaring both sides, we get

$4 λ^{2} - 9 > \frac{25}{16} \Rightarrow 4 λ^{2} > \frac{169}{16} \Rightarrow λ^{2} > \frac{169}{64}$

$\Rightarrow λ > \frac{13}{8} or λ < - \frac{13}{8}$

$∴ λ \in (- \infty, - \frac{13}{8}) \cup (\frac{13}{8}, \infty)$

Also, $4 λ^{2} < {(2 + \sqrt{4 λ^{2} - 9})}^{2}$ [By (iii)]

$\Rightarrow 4 λ^{2} < 4 + 4 \sqrt{4 λ^{2} - 9} + 4 λ^{2} - 9$

$\Rightarrow 0 < - 5 + 4 \sqrt{4 λ^{2} - 9} \Rightarrow 5 < 4 \sqrt{4 λ^{2} - 9}$

On squaring, we get

$\frac{25}{16} < 4 λ^{2} - 9 \Rightarrow \frac{169}{64} < λ^{2}$

$∴ λ \in (- \infty, - \frac{13}{8}) \cup (\frac{13}{8}, \infty)$

Thus, Circles C and C' intersect at two distinct points for

$λ \in R - [\frac{- 13}{8}, \frac{13}{8}]$

$∴ a = \frac{- 13}{8}, b = \frac{13}{8}$

$∴$ (8a + 12, 16b –20) = (–1, 6) which satisfies only $6 x^{2} + y^{2} = 42$ .

Q 10 :

Let the locus of the midpoints of the chords of the circle $x^{2} + {(y - 1)}^{2} = 1$ drawn from the origin intersect the line x + y = 1 at P and Q. Then, the length of PQ is : [2024]

$\frac{1}{2}$
1
$\frac{1}{\sqrt{2}}$
$\sqrt{2}$

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(3)

Let $(x_{1}, y_{1})$ be the mid point of chords.

So, equation of chord of the circle $x^{2} + y^{2} - 2 y = 0$ is,

$x \cdot x_{1} + y \cdot y_{1} - (y + y_{1}) = x_{1}^{2} + y_{1}^{2} - 2 y_{1}$

The chord is passing through origin

$∴ - y_{1} = x_{1}^{2} + y_{1}^{2} - 2 y_{1}$

$x_{1}^{2} + y_{1}^{2} - y_{1} = 0$ ... (i)

Now, (i) intersects the line x + y = 1

$∴ {(1 - y_{1})}^{2} + y_{1}^{2} - y_{1} = 0$

$\Rightarrow 1 + y_{1}^{2} - 2 y_{1} + y_{1}^{2} - y_{1} = 0 \Rightarrow 2 y_{1}^{2} - 3 y_{1} + 1 = 0$

$\Rightarrow (2 y_{1} - 1) (y_{1} - 1) = 0 \Rightarrow y_{1} = \frac{1}{2} or y_{1} = 1$

If $y_{1} = \frac{1}{2}$ , then $x_{1} = \frac{1}{2}$ [From (i)]

If y = 1, then $x_{1} = 0$

So, $P = (\frac{1}{2}, \frac{1}{2})$ and Q = (1, 0) $∴ P Q = \sqrt{{(\frac{1}{2})}^{2} + {(\frac{- 1}{2})}^{2}} = \frac{1}{\sqrt{2}}$

Q 11 :

Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to : [2024]

$\frac{2}{13}$
$\frac{5}{13}$
$\frac{3}{13}$
$\frac{1}{13}$

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(2)

Let the equation of circle be

${(x - x_{1})}^{2} + {(y - y_{1})}^{2} = r^{2}$ ...(i)

Put x = 0, y = 0 in (i), we get

$\Rightarrow x_{1}^{2} + y_{1}^{2} = r^{2}$ ... (ii)

Put x = 0, y = 1 in (i), we get

$\Rightarrow x_{1}^{2} + {(1 - y_{1})}^{2} = r^{2}$ ... (iii)

From (ii) and (iii), we get

$y_{1}^{2} = {(1 - y_{1})}^{2} \Rightarrow y_{1}^{2} = 1 + y_{1}^{2} - 2 y_{1} \Rightarrow y_{1} = \frac{1}{2}$

Put x = 1, y = 0 in (i), we get

${(1 - x_{1})}^{2} + y_{1}^{2} = r^{2}$ ... (iv)

From (ii) and (iv), we get

${(1 - x_{1})}^{2} = x_{1}^{2} \Rightarrow 1 + x_{1}^{2} - 2 x_{1} = x_{1}^{2} \Rightarrow x_{1} = \frac{1}{2}$

$∴$ From (ii), we get $r = \frac{1}{\sqrt{2}}$

Putting $x_{1} = y_{1} = \frac{1}{2}$ and $r = \frac{1}{\sqrt{2}}$ in (i), we get

${(x - \frac{1}{2})}^{2} + {(y - \frac{1}{2})}^{2} = \frac{1}{2}$

Point (2k, 3k) also satisfies the equation of circle.

${(2 k - \frac{1}{2})}^{2} + {(3 k - \frac{1}{2})}^{2} = \frac{1}{2}$

$\Rightarrow 4 k^{2} + \frac{1}{4} - 2 k + 9 k^{2} + \frac{1}{4} - 3 k = \frac{1}{2}$

$\Rightarrow 13 k^{2} = 5 k \Rightarrow k = \frac{5}{13}$

Q 12 :

If the circles ${(x + 1)}^{2} + {(y + 2)}^{2} = r^{2}$ and $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ intersect at exactly two distinct points, then [2024]

5 < r < 9
3 < r < 7
0 < r < 7
$\frac{1}{2} < r < 7$

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(2)

Let $P : {(x + 1)}^{2} + {(y + 2)}^{2} = r^{2}$ and $Q : x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ be two given circles.

Q can be written as ${(x - 2)}^{2} + {(y - 2)}^{2} = 4$

Centre of circle P and Q are (–1, –2) and (2, 2) respectively

Distance between centre of circle is given by

$D = \sqrt{{(2 + 1)}^{2} + {(2 + 2)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 units$

For the intersection of circles, $D > | r_{P} - r_{Q} |$ and $D < (r_{P} + r_{Q})$ , where $r_{P}$ and $r_{Q}$ are radius of circle P and Q respectively

$\Rightarrow 5 > | r - 2 |$ and 5 < r + 2

$\Rightarrow - 5 < r - 2 < 5 \Rightarrow - 3 < r < 7$ ... (i)

and r > 3 ... (ii)

From (i) and (ii), 3 < r < 7.

Q 13 :

If one of the diameters of the circle $x^{2} + y^{2} - 10 x + 4 y + 13 = 0$ is a chord of another circle C, whose centre is the point of intersection of the lines 2x + 3y = 12 and 3x –2y = 5, then the radius of the circle C is : [2024]

6
4
$\sqrt{20}$
$3 \sqrt{2}$

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(1)

Given, 2x + 3y = 12

3x – 2y = 5

Point of intersection = (3, 2)

$∴$ Centre is (3, 2)

$l = \sqrt{4 + 16} = 2 \sqrt{5}$

$l^{2} + R^{2} = r^{2}$

$\Rightarrow 20 + (25 + 4 - 13) = r^{2}$

$\Rightarrow 20 + 16 = r^{2} \Rightarrow r = 6$

Q 14 :

Let a variable line passing through the centre of the circle $x^{2} + y^{2} - 16 x - 4 y = 0$ , meet the positive co-ordinate axes at the points A and B. Then the minimum value of OA + OB, where O is the origin is equal to [2024]

18
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(1)

Given circle is $x^{2} + y^{2} - 16 x - 4 y = 0$

Centre is (– g, – f)

Now, 2g = 16 $\Rightarrow$ g = 8 and 2f = 4 $\Rightarrow$ f = 2

$∴$ Centre is (8, 2)

Equation of line $l_{1}$ is ... (i)

y – 2 = m(x – 8)

Equation (i) cuts the x-axis then y = 0

$\frac{- 2}{m} = x - 8 \Rightarrow x = \frac{8 m - 2}{m} = O A$

Equation (i) cuts the y-axis, then x = 0

y – 2 = – 8m $\Rightarrow$ y = 2 – 8m = OB

Let $l = O A + O B = \frac{8 m - 2}{m} + 2 - 8 m = 10 - 8 m - \frac{2}{m}$ ... (ii)

Differentiate (ii) w.r.t. m we get

$\frac{d l}{d m} = - 8 + \frac{2}{m^{2}}$ ... (iii)

$\frac{d l}{d m} = 0 \Rightarrow - 8 + \frac{2}{m^{2}} = 0 \Rightarrow m = \pm \frac{1}{2}$

Differentiate (iii), w.r.t. m, we get

$\frac{d^{2} l}{d m^{2}} = \frac{- 4}{m^{3}}$

Now, $\frac{d^{2} l}{d m^{2}}$ $|_{m = \frac{- 1}{2}} = \frac{- 4}{{(\frac{- 1}{2})}^{3}} = 32 a n d \frac{d^{2} l}{d m^{2}} |_{m = \frac{1}{2}}$ $= \frac{- 4}{{(\frac{1}{2})}^{3}} = - 32$

$∴ \frac{d^{2} l}{d m^{2}} > 0 at m = \frac{- 1}{2}$

Hence, minima occurs at $m = \frac{- 1}{2}$

So, the minimum value of $l$ is

$= 10 - 8 \times (- \frac{1}{2}) - 2 (- 2) = 10 + 4 + 4 = 18$ .

Q 15 :

Let the maximum and minimum values of ${(\sqrt{8 x - x^{2} - 12} - 4)}^{2} + {(x - 7)}^{2}, x \in R$ be M and m, respectively. Then $M^{2} - m^{2}$ is equal to __________. [2024]

(1600)

Let $y = \sqrt{8 x - x^{2} - 12}$

$\Rightarrow y^{2} = 8 x - x^{2} - 12$

$= - (x^{2} - 8 x + 16) + 4 = - {(x - 4)}^{2} + 4$

$\Rightarrow {(x - 4)}^{2} + y^{2} = 4$ ... (i)

which is a circle with centre (4, 0) and radius 2.

Now, ${(y - 4)}^{2} + {(x - 7)}^{2}$ represent distance of (x, y) from (7, 4)

M = Maxium distance = ${(2 - 7)}^{2} + {(0 - 4)}^{2} = 25 + 16 = 41$

m = Minimum distance = Distance between P and (7, 4)

where P is the intersection of circle with line joining (4, 0) and (7, 4).

Now, equation of line joining (4, 0) and (7, 4) is given by

$(y - 0) = \frac{4 - 0}{7 - 4} (x - 4)$

i.e., $y = \frac{4}{3} (x - 4)$

On substituting the value of y is in (i), we get

${(x - 4)}^{2} + \frac{16}{9} {(x - 4)}^{2} = 4$

$\Rightarrow {(x - 4)}^{2} [\frac{25}{9}] = 4 \Rightarrow {(x - 4)}^{2} = \frac{36}{25} \Rightarrow x - 4 = \pm \frac{6}{5}$

$\Rightarrow x = \frac{6}{5} + 4 = \frac{26}{5}, y = \frac{8}{5}$ (Neglect negative sign)

So, $m = {(\frac{26}{5} - 7)}^{2} + {(\frac{8}{5} - 4)}^{2} = \frac{81}{25} + \frac{144}{25} = \frac{225}{25} = 9$

$∴ M^{2} - m^{2} = 41^{2} - 9^{2} = 1600$ .

Q 16 :

Let the centre of a circle, passing through the points (0, 0), (1, 0) and touching the circle $x^{2} + y^{2} = 9$ , be (h, k). Then for all possible values of the coordinates of the centre (h, k), $4 (h^{2} + k^{2})$ is equal to _________. [2024]

(9)

Circle will touch internally

$∴ C_{1} C_{2} = | r_{1} - r_{2} |$

$\Rightarrow \sqrt{h^{2} + k^{2}} = 3 - \sqrt{h^{2} + k^{2}} \Rightarrow 2 \sqrt{h^{2} + k^{2}} = 3$

$\Rightarrow h^{2} + k^{2} = \frac{9}{4} ∴ 4 (h^{2} + k^{2}) = 9$ .

Q 17 :

Consider a circle ${(x - α)}^{2} + {(y - β)}^{2} = 50$ , where $α, β > 0$ . If the circle touches the line y + x = 0 at the point P, whose distance from the origin is $4 \sqrt{2}$ , then ${(α + β)}^{2}$ is equal to __________. [2024]

(100)

We have circle,

${(x - α)}^{2} + {(y - β)}^{2} = 50$

Centre of (i), is $C (α, β)$ and radius, $r = 5 \sqrt{2}$

$∴ C P = r$

$\Rightarrow$ $| \frac{α + β}{\sqrt{2}} |$ $= 5 \sqrt{2}$

$\Rightarrow | α + β | = 10 \Rightarrow {(α + β)}^{2} = 100$ .

Q 18 :

Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to __________. [2024]

(2)

Solving 2x – 3y = 5 and 3x – 4y = 7, we get x = 1 and y = –1

Now, equation of tangent joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ is

$y + 4 = \frac{3 + 4}{\frac{- 1}{7} + \frac{22}{7}} (x + \frac{22}{7}) \Rightarrow 3 (y + 4) = 7 (x + \frac{22}{7})$

$\Rightarrow 7 x - 3 y + 10 = 0$

$α, β$ be the foot of perpendicular from point (1, –1) to the line 7x – 3y + 10 = 0.

$∴ \frac{α - 1}{7} = \frac{β + 1}{- 3} = \frac{- (7 (1) - 3 (- 1) + 10)}{{(- 3)}^{2} + 7^{2}} = - \frac{20}{58} = \frac{- 10}{29}$

$∴ α = \frac{- 70}{29} + 1 = \frac{- 41}{29}, β = \frac{30}{29} - 1 = \frac{1}{29}$

$∴ 17 β - α = \frac{17}{29} + \frac{41}{29} = \frac{58}{29} = 2$ .

Q 19 :

Consider two circles $C_{1} : x^{2} + y^{2} = 25$ and $C_{2} : {(x - α)}^{2} + y^{2} = 16$ , where $α \in (5, 9)$ . Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $C_{1}$ and $C_{2}$ be $\sin^{- 1} (\frac{\sqrt{63}}{8})$ . If the length of common chord of $C_{1}$ and $C_{2}$ is $β$ , then the value of ${(α β)}^{2}$ equals __________. [2024]

(1575)

We have,

$C_{1} : x^{2} + y^{2} = 25$ or $C_{2} : {(x - α)}^{2} + y^{2} = 16$

Let $θ$ be the angle between two radii

$∴ θ = \sin^{- 1} \frac{\sqrt{63}}{8}$

$\Rightarrow \sin θ = \frac{\sqrt{63}}{8}$

Area of $△ O A B = \frac{1}{2} \times 4 \times 5 \sin θ$

$\Rightarrow \frac{1}{2} \times α \times \frac{β}{2} = 10 \frac{\sqrt{63}}{8} \Rightarrow α β = 5 \sqrt{63} \Rightarrow {(α β)}^{2} = 1575$ .

Q 20 :

If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then $10 k + r^{2}$ is equal to [2025]

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33
32
35

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2

3

4

(4)

Let $P \equiv (4, 6), Q \equiv (- 1, 5), R \equiv (0, 0)$ and $S \equiv (k, 3 k)$ .

Slope of line through point P and Q is

$m_{1} = \frac{1}{5}$ ... (i)

Slope of line through point Q and R is

$m_{2} = - 5$ ... (ii)

From (i) and (ii), we get

$m_{1} m_{2} = - 1$

So, line passes by P and Q is perpendicular to line passes by Q and R.

So, P and R will represent end point of diameter of circle so we have

(x – 4)(x – 0) + (y – 6) (y – 0) = 0

$\Rightarrow x^{2} + y^{2} - 4 x - 6 y = 0$

Since (k, 3k) lies on it

$∴ k^{2} + 9 k^{2} - 4 k - 18 k = 0$

$\Rightarrow 10 k^{2} - 22 k = 0$

$\Rightarrow$ k = 0, $\frac{11}{5}$

Since, k = 0 is not possible, so $k = \frac{11}{5}$

Also, $r = \frac{\sqrt{4^{2} + 6^{2}}}{2} = \frac{\sqrt{52}}{2} = \sqrt{13}$

$∴ 10 k + r^{2} = 10 \times \frac{11}{5} + {(\sqrt{13})}^{2} = 35$ .

Q 21 :

Let $C_{1}$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $C_{2}$ be the circle with centre (1, 3) that touches $C_{1}$ externally at the point $(α, β)$ . If ${(β - α)}^{2} = \frac{m}{n}$ , gcd (m, n) = 1, then m + n is equal to [2025]

31
13
9
22

1

2

3

4

(4)

The equation of circle $C_{1}, {(x + 3)}^{2} + {(y + 3)}^{2} = 3^{2}$

The centres has $C_{1}$ and $C_{2}$ are A(–3, –3) and B(1, 3)

$A B = \sqrt{16 + 36} = 2 \sqrt{13}$

So, the radius are $r_{1} = 3$ and $r_{2} = 2 \sqrt{13} - 3$

The point $P (α, β)$

$α = \frac{r_{1} (1) + r_{2} (- 3)}{r_{1} + r_{2}}, β = \frac{r_{1} (3) + r_{2} (- 3)}{r_{1} + r_{2}}$

$α = \frac{3 - 3 (2 \sqrt{13} - 3)}{2 \sqrt{13}}, β = \frac{3 (3) + (2 \sqrt{3} - 3) (- 3)}{2 \sqrt{13}}$

$\Rightarrow α = \frac{12 - 6 \sqrt{3}}{2 \sqrt{3}}, β = \frac{18 - 6 \sqrt{3}}{2 \sqrt{3}}$

Now, ${(β - α)}^{2} = {(\frac{6}{2 \sqrt{13}})}^{2} = \frac{36}{52} = \frac{9}{13}$ $(Given, {(β - α)}^{2} = \frac{m}{n})$

So, m = 9, n = 13.

So, m + n = 22.

Q 22 :

A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point (2, 5) and intersects the circle C at exactly two points. If the set of all possible values of r is the interval $(α, β)$ , then $3 β - 2 α$ is equal to: [2025]

12
14
10
15

1

2

3

4

(4)

Circle with radius r touches the circle C, when r + 2 = distance between their centres

i.e., $r + 2 = \sqrt{4^{2} + 3^{2}} = 5$

Also, if circle C touches the circle with radius r internally, then

$r = 2 + \sqrt{4^{2} + 3^{2}} = 2 + 5 = 7$

Since, circle with radius r intersects the circle C at exactly 2 points.

$∴ r + 2 > 5 and r < 7 i . e ., 3 < r < 7$

$∴ α = 3, β = 7$

$\Rightarrow 3 β - 2 α = (3) (7) - (2) (3) = 21 - 6 = 15$ .

Q 23 :

Let circle C be the image of $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ in the line 2x – 3y + 5 = 0 and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C. If $B (α, β)$ , with $β < 4$ , lies on C such that the length of the arc AB is ${(1 / 6)}^{th}$ of the perimeter of C, then $β - \sqrt{3} α$ is equal to [2025]

$3 + \sqrt{3}$
$4 - \sqrt{3}$
4
3

1

2

3

4

(3)

Centre of the circle $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ is (1, –2) and its radius $\sqrt{1 + 4 + 4} = 3$

Image of (1, –2) about 2x – 3y + 5 = 0 is

$\frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{- 2 (2 + 6 + 5)}{4 + 9} = - 2$

$∴$ x = – 3 and y = 4

Centre of circle C is (–3, 4).

Equation of Circle C is ${(x + 3)}^{2} + {(y - 4)}^{2} = 9$

$l$ $(arc A B) = \frac{1}{6} \times 2 π r \Rightarrow r θ = \frac{1}{6} \times 2 π r \Rightarrow θ = \frac{π}{3}$

As $(α, β)$ lies on circle C,

$\Rightarrow {(α + 3)}^{2} + {(β - 4)}^{2} = 9$ ... (i)

$\Rightarrow$ Coordinates of A are (0, 4) ( $∵$ A lies on right side of O and OA $| |$ x-axis)

Now, $A B^{2} = O A^{2} + O B^{2} - 2 O A \cdot O B \cdot \cos \frac{π}{3}$

$= 9 + 9 - 2 \times 3 \times 3 \times \frac{1}{2} = 9$

$\Rightarrow α^{2} + {(β - 4)}^{2} = 9$ ... (ii)

From (i) and (ii), we get $α = \frac{- 3}{2}$

${(β - 4)}^{2} = 9 - \frac{9}{4} \Rightarrow β = \frac{- 3 \sqrt{3}}{2} + 4 (∵ β < 4)$

Now, $β - \sqrt{3} α = \frac{- 3 \sqrt{3}}{2} + 4 - \sqrt{3} (\frac{- 3}{2}) = 4$ .

Q 24 :

Let the equation of the circle, which touches x-axis at the point (a, 0), a > 0 and cuts off an intercept of length b on y-axis be $x^{2} + y^{2} - α x + β y + γ = 0$ . If the circle lies below x-axis, then the ordered pair ( $2 a, b^{2}$ ) is equal to [2025]

$(γ, β^{2} + 4 α)$
$(γ, β^{2} - 4 α)$
$(α, β^{2} + 4 γ)$
$(α, β^{2} - 4 γ)$

1

2

3

4

(4)

Let, r be the radius of the circle,

$r = \sqrt{\frac{α^{2}}{4} + \frac{β^{2}}{4} - γ} = \frac{β}{2}$

$\Rightarrow \frac{α^{2}}{4} - γ = 0$

$\Rightarrow α^{2} = 4 γ; \frac{α}{2} = a \Rightarrow α = 2 a$

Now, length of intercept on y-axis = $b = 2 \sqrt{\frac{β^{2}}{4} - γ}$

$\Rightarrow \frac{β^{2}}{4} - γ = \frac{b^{2}}{4} \Rightarrow b^{2} = β^{2} - 4 γ$

$∴$ Points ( $2 a, b^{2}$ ) = $(α, β^{2} - 4 γ)$

Q 25 :

Let the line x + y = 1 meet the circle $x^{2} + y^{2} = 4$ at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to: [2025]

$3 \sqrt{7}$
$\sqrt{14}$
$2 \sqrt{14}$
$5 \sqrt{7}$

1

2

3

4

(3)

For points of intersection A and B.

Solving x + y = 1 and $x^{2} + y^{2} = 4$ , we get

$A (\frac{1 - \sqrt{7}}{2}, \frac{1 + \sqrt{7}}{2}), B (\frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2})$

Slope of AB = –1

Slope of $⊥^{r}$ bisector of AB = 1

For points of intersection C and D

Solving, x = y and $x^{2} + y^{2} = 4$ , we get

$C (\sqrt{2}, \sqrt{2})$ and $D (- \sqrt{2}, - \sqrt{2})$

$∴$ Area of quadrilateral ADBC = 2 $\times$ Area of $△$ BCD

$= 2 \times \frac{1}{2}$ $| \begin{matrix} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1 + \sqrt{7}}{2} & \frac{1 - \sqrt{7}}{2} & 1 \\ - \sqrt{2} & - \sqrt{2} & 1 \end{matrix} |$ $= 2 \sqrt{14}$ sq. units.

Q 26 :

Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C, whose mid-point is (1, 2), is: [2025]

$2 \sqrt{2}$
$2 \sqrt{3}$
$4 \sqrt{2}$
$\sqrt{3}$

1

2

3

4

(2)

Since, centre (h, k) lies on 3x + 2y + 2 = 0

$\Rightarrow$ 3h + 2k + 2 = 0 ... (i)

Also, the circle passes through the points (4, 2) and (0, 2), then we can say that ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ passes through (4, 2) and (0, 2)

$∴ h^{2} + {(2 - k)}^{2} = r^{2}$ ... (ii)

and ${(4 - h)}^{2} + {(2 - k)}^{2} = r^{2}$ ... (iii)

On subtracting (ii) from (iii), we get

$h^{2} = {(4 - h)}^{2} \Rightarrow h^{2} = 16 + h^{2} - 8 h \Rightarrow h = 2$

From (i), k = – 4

$∴$ Centre = (2, –4)

Radius, $r = \sqrt{{(0 - 2)}^{2} + {(2 + 4)}^{2}} = \sqrt{40}$

Now, mid-point of the chord is (1, 2)

$\Rightarrow$ Perpendicular distance from centre to chord = d

$= \sqrt{{(2 - 1)}^{2} + {(- 4 - 2)}^{2}} = \sqrt{37}$

$∴$ Length of chord = $2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{40 - 37} = 2 \sqrt{3}$

Q 27 :

The absolute difference between the squares of the radii of the two circles passing through the point (–9, 4) and touching the lines x + y = 3 and x – y = 3, is equal to __________. [2025]

(768)

We have, x + y = 3 and x – y = 3 are tangents

$∴$ The centre of both circles will lie on x-axis

Equation of circle is

$∴ {(x - α)}^{2} + y^{2} = r^{2}$

Hence, centre is C( $α$ , 0).

$r = P C = \sqrt{{(α + 9)}^{2} + 16}$ ... (i)

Also, $| \frac{α - 3}{\sqrt{2}} |$ $= r$ ... (ii)

From (i) and (ii), we get

$\sqrt{{(α + 9)}^{2} + 16} =$ $| \frac{α - 3}{\sqrt{2}} |$

$\Rightarrow α = - 5 or - 37$

$r =$ $| \frac{- 5 - 3}{\sqrt{2}} |$ $or$ $| \frac{- 37 - 3}{\sqrt{2}} |$ $= 4 \sqrt{2} or 20 \sqrt{2}$

$∴$ Now, $| r_{1}^{2} - r_{2}^{2} |$ $=$ $| {(4 \sqrt{2})}^{2} - {(20 \sqrt{2})}^{2} |$ $= 768$ .

Q 28 :

Let r be the radius of the circle, which touches x-axis at point (a, 0), a < 0 and the parabola $y^{2} = 9 x$ at the point (4, 6). Then r is equal to __________. [2025]

(30)

Equation of circle is given by

${(x - a)}^{2} + {(y - r)}^{2} = r^{2}$

Now, circle passes through the point A(4, 6). So, we have

${(4 - a)}^{2} + {(6 - r)}^{2} = r^{2}$

$\Rightarrow 16 + a^{2} - 8 a + 36 + r^{2} - 12 r = r^{2}$

$\Rightarrow a^{2} - 12 r - 8 a + 52 = 0$ ... (i)

Equation of tangent to the barabola $y^{2} = 9 x$ at the point A(4, 6) is given by $y y_{1} = 2 a (x + x_{1})$

$\Rightarrow 6 y = \frac{2 \times 9}{4} (x + 4) \Rightarrow 6 y = \frac{9 x + 36}{2}$

$\Rightarrow 3 x - 4 y + 12 = 0$

Distance of this line from centre of circle is equal to radius of circle.

$∴$ $| \frac{3 a - 4 r + 12}{\sqrt{9 + 16}} |$ $= r$

$\Rightarrow 3 a - 4 r + 12 = \pm 5 r$

$\Rightarrow 3 a + 12 = 9 r or 3 a + 12 = - r$

If 3a + 12 = 9r i.e., a + 4 = 3r, then by using equation (i), we get

$a^{2} - 12 (\frac{a + 4}{3}) - 8 a + 52 = 0$

$\Rightarrow a^{2} - 4 a - 16 - 8 a + 52 = 0 \Rightarrow a^{2} - 12 a + 36 = 0$

$\Rightarrow {(a - 6)}^{2} = 0 \Rightarrow a = 6 but a < 0 \Rightarrow a \neq 6$

Now, if 3a + 12 = – r, so by using equation (i), we get

$a^{2} - 8 a + 12 (3 a + 12) + 52 = 0$

$\Rightarrow a^{2} + 28 a + 196 = 0 \Rightarrow {(a + 14)}^{2} = 0$

$\Rightarrow a = - 14 \Rightarrow r = 30$ .

Q 29 :

If the tangents at the point P and Q on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle PQR is [2023]

$\frac{13}{4}$
$\frac{5}{4}$
$\frac{5}{8}$
$\frac{13}{8}$

1

2

3

4

(3)

Equation of circle(s) is, $x^{2} + y^{2} - 2 x + y - 5 = 0$

$\Rightarrow {(x - 1)}^{2} + {(y + \frac{1}{2})}^{2} = {(\frac{5}{2})}^{2}$

$\Rightarrow Radius (R) = \frac{5}{2}$

Now, length of tangent, $L = \sqrt{S_{1}} = \sqrt{\frac{81}{16} + 4 - 2 \times \frac{9}{4} + 2 - 5} = \frac{5}{4}$

So, area of $∆ P Q R$ = $\frac{R L^{3}}{R^{2} + L^{2}} = \frac{(\frac{5}{2}) {(\frac{5}{4})}^{3}}{{(\frac{5}{2})}^{2} + {(\frac{5}{4})}^{2}} = \frac{5}{8}$

Q 30 :

Let O be the origin and OP and OQ be the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point $(α, \frac{1}{2}),$ then a value of $α$ is [2023]

$\frac{5}{2}$
$\frac{3}{2}$
$- \frac{1}{2}$
$1$

1

2

3

4

(1)

PQ is the chord of contact of the tangents from the origin to the circle, $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ ...(i)

Equation of $P Q$ is, $3 x - 2 y - 8 = 0$ ...(ii)

Equation of circle passing through the intersection of (i) and (ii) is, $x^{2} + y^{2} - 6 x + 4 y + 8 + λ (3 x - 2 y - 8) = 0$ ...(iii)

If this represents the circumcircle of the triangle $O P Q$ , it passes through $O (0, 0)$ .

So, $λ = 1$ and (iii) becomes

$x^{2} + y^{2} - 3 x + 2 y = 0$ ...(iv)

Given, $(α, \frac{1}{2})$ passes through (iv)

$∴$ $α^{2} + \frac{1}{4} - 3 α + 1 = 0$ $\Rightarrow 4 α^{2} - 12 α + 5 = 0$

$\Rightarrow (2 α - 1) (2 α - 5) = 0$ $\Rightarrow α = \frac{1}{2} or \frac{5}{2}$

Hence, the value of $α$ is $\frac{5}{2}$ .

Topic Question Set

Q 1 :

A square is inscribed in the circle $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$ . One side of this square is parallel to y = x + 3. If $(x_{i}, y_{i})$ are the vertices of the square, then $\sum_{}^{} (x_{i}^{2}, y_{i}^{2})$ is equal to [2024]

Q 2 :

Let C be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line x + y = 2 intersects the circle C at the points P and Q. Let MN be a chord of C of length 2 units and slope –1. Then a distance (in units) between the chord PQ and the chord MN is [2024]

Q 3 :

Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3, 2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5, 5) is: [2024]

Q 4 :

Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at (–1, 0) and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the y-axis at the point P, then the square of the distance of P from the centre of $C_{1}$ is: [2024]

Q 5 :

Let ABCD and AEFG be squares of sides 4 and 2 units respectively. the point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies : [2024]

Q 7 :

If the image of the point (–4,5) in the line x + 2y = 2 lies on the circle ${(x + 4)}^{2} + {(y - 3)}^{2} = r^{2}$ , then r is equal to : [2024]

Q 8 :

Let a circle passing through (2,0) have its centre at the point (h, k). Let $(x_{c}, y_{c})$ be the point of intersection of the lines 3x + 5y = 1 and $(2 + c) x + 5 c^{2} y = 1$ . If $h = \lim_{c \to 1} x_{c}$ and $k = \lim_{c \to 1} y_{c}$ , then the equation of the circle is : [2024]

Q 9 :

Let $C : x^{2} + y^{2} = 4$ and $C' : x^{2} + y^{2} - 4 λ x + 9 = 0$ be two circles. If the set of all values of $λ$ so that the circles C and C' intersect at two distinct points, is R – [a, b], then the point (8a + 12, 16b – 20) lies on the curve : [2024]

Q 10 :

Let the locus of the midpoints of the chords of the circle $x^{2} + {(y - 1)}^{2} = 1$ drawn from the origin intersect the line x + y = 1 at P and Q. Then, the length of PQ is : [2024]

Q 11 :

Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to : [2024]

Q 12 :

If the circles ${(x + 1)}^{2} + {(y + 2)}^{2} = r^{2}$ and $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ intersect at exactly two distinct points, then [2024]

Q 13 :

If one of the diameters of the circle $x^{2} + y^{2} - 10 x + 4 y + 13 = 0$ is a chord of another circle C, whose centre is the point of intersection of the lines 2x + 3y = 12 and 3x –2y = 5, then the radius of the circle C is : [2024]

Q 14 :

Let a variable line passing through the centre of the circle $x^{2} + y^{2} - 16 x - 4 y = 0$ , meet the positive co-ordinate axes at the points A and B. Then the minimum value of OA + OB, where O is the origin is equal to [2024]

Q 15 :

Let the maximum and minimum values of ${(\sqrt{8 x - x^{2} - 12} - 4)}^{2} + {(x - 7)}^{2}, x \in R$ be M and m, respectively. Then $M^{2} - m^{2}$ is equal to __________. [2024]

Q 16 :

Let the centre of a circle, passing through the points (0, 0), (1, 0) and touching the circle $x^{2} + y^{2} = 9$ , be (h, k). Then for all possible values of the coordinates of the centre (h, k), $4 (h^{2} + k^{2})$ is equal to _________. [2024]

Q 17 :

Consider a circle ${(x - α)}^{2} + {(y - β)}^{2} = 50$ , where $α, β > 0$ . If the circle touches the line y + x = 0 at the point P, whose distance from the origin is $4 \sqrt{2}$ , then ${(α + β)}^{2}$ is equal to __________. [2024]

Q 18 :

Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to __________. [2024]

Q 20 :

If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then $10 k + r^{2}$ is equal to [2025]

Q 21 :

Let $C_{1}$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $C_{2}$ be the circle with centre (1, 3) that touches $C_{1}$ externally at the point $(α, β)$ . If ${(β - α)}^{2} = \frac{m}{n}$ , gcd (m, n) = 1, then m + n is equal to [2025]

Q 24 :

Let the equation of the circle, which touches x-axis at the point (a, 0), a > 0 and cuts off an intercept of length b on y-axis be $x^{2} + y^{2} - α x + β y + γ = 0$ . If the circle lies below x-axis, then the ordered pair ( $2 a, b^{2}$ ) is equal to [2025]

Q 25 :

Let the line x + y = 1 meet the circle $x^{2} + y^{2} = 4$ at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to: [2025]

Q 26 :

Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C, whose mid-point is (1, 2), is: [2025]

Q 27 :

The absolute difference between the squares of the radii of the two circles passing through the point (–9, 4) and touching the lines x + y = 3 and x – y = 3, is equal to __________. [2025]

Q 28 :

Let r be the radius of the circle, which touches x-axis at point (a, 0), a < 0 and the parabola $y^{2} = 9 x$ at the point (4, 6). Then r is equal to __________. [2025]

Q 29 :

If the tangents at the point P and Q on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle PQR is [2023]

Q 30 :

Let O be the origin and OP and OQ be the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point $(α, \frac{1}{2}),$ then a value of $α$ is [2023]

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Topic Question Set

Q 1 : A square is inscribed in the circle x2+y2–10x–6y+30=0. One side of this square is parallel to y = x + 3. If (xi, yi) are the vertices of the square, then ∑(xi2, yi2) is equal to [2024]

Q 2 : Let C be a circle with radius 10 units and centre at the origin. Let the line x + y = 2 intersects the circle C at the points P and Q. Let MN be a chord of C of length 2 units and slope –1. Then a distance (in units) between the chord PQ and the chord MN is [2024]

Q 3 : Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3, 2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5, 5) is: [2024]

Q 4 : Let the circle C1:x2+y2–2(x+y)+1=0 and C2 be a circle having centre at (–1, 0) and radius 2. If the line of the common chord of C1 and C2 intersects the y-axis at the point P, then the square of the distance of P from the centre of C1 is: [2024]

Q 5 : Let ABCD and AEFG be squares of sides 4 and 2 units respectively. the point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies : [2024]

Q 6 : Let the circles C1:(x–α)2+(y–β)2=r12 and C2:(x–8)2+(y–152)2=r22 touch each other externally at the point (6, 6). If the point (6, 6) divides the line segment joining the centres of the circles C1 and C2 internally in the ratio 2 : 1, then (α+β)+4(r12+r22) equals [2024]

Q 7 : If the image of the point (–4,5) in the line x + 2y = 2 lies on the circle (x+4)2+(y–3)2=r2, then r is equal to : [2024]

Q 8 : Let a circle passing through (2,0) have its centre at the point (h, k). Let (xc,yc) be the point of intersection of the lines 3x + 5y = 1 and (2+c)x+5c2y=1. If h=limc→1xc and k=limc→1yc, then the equation of the circle is : [2024]

Q 9 : Let C:x2+y2=4 and C':x2+y2–4λx+9=0 be two circles. If the set of all values of λ so that the circles C and C' intersect at two distinct points, is R – [a, b], then the point (8a + 12, 16b – 20) lies on the curve : [2024]

Q 10 : Let the locus of the midpoints of the chords of the circle x2+(y–1)2=1 drawn from the origin intersect the line x + y = 1 at P and Q. Then, the length of PQ is : [2024]

Q 11 : Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to : [2024]

Q 12 : If the circles (x+1)2+(y+2)2=r2 and x2+y2–4x–4y+4=0 intersect at exactly two distinct points, then [2024]

Q 13 : If one of the diameters of the circle x2+y2–10x+4y+13=0 is a chord of another circle C, whose centre is the point of intersection of the lines 2x + 3y = 12 and 3x –2y = 5, then the radius of the circle C is : [2024]

Q 14 : Let a variable line passing through the centre of the circle x2+y2–16x–4y=0, meet the positive co-ordinate axes at the points A and B. Then the minimum value of OA + OB, where O is the origin is equal to [2024]

Q 15 : Let the maximum and minimum values of (8x–x2–12–4)2+(x-7)2, x∈R be M and m, respectively. Then M2–m2 is equal to __________. [2024]

Q 16 : Let the centre of a circle, passing through the points (0, 0), (1, 0) and touching the circle x2+y2=9, be (h, k). Then for all possible values of the coordinates of the centre (h, k), 4(h2+k2) is equal to _________. [2024]

Q 17 : Consider a circle (x–α)2+(y–β)2=50, where α,β>0. If the circle touches the line y + x = 0 at the point P, whose distance from the origin is 42, then (α+β)2 is equal to __________. [2024]

Q 18 : Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points (–227,–4) and (–17,3) intersects the circle at only one point P(α,β). Then 17β–α is equal to __________. [2024]

Q 20 : If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then 10k+r2 is equal to [2025]

Q 21 : Let C1 be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let C2 be the circle with centre (1, 3) that touches C1 externally at the point (α,β). If (β–α)2=mn, gcd (m, n) = 1, then m + n is equal to [2025]

Q 24 : Let the equation of the circle, which touches x-axis at the point (a, 0), a > 0 and cuts off an intercept of length b on y-axis be x2+y2–αx+βy+γ=0. If the circle lies below x-axis, then the ordered pair (2a,b2) is equal to [2025]

Q 25 : Let the line x + y = 1 meet the circle x2+y2=4 at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to: [2025]

Q 26 : Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C, whose mid-point is (1, 2), is: [2025]

Q 27 : The absolute difference between the squares of the radii of the two circles passing through the point (–9, 4) and touching the lines x + y = 3 and x – y = 3, is equal to __________. [2025]

Q 28 : Let r be the radius of the circle, which touches x-axis at point (a, 0), a < 0 and the parabola y2=9x at the point (4, 6). Then r is equal to __________. [2025]

Q 29 : If the tangents at the point P and Q on the circle x2+y2-2x+y=5 meet at the point R(94,2), then the area of the triangle PQR is [2023]

Q 30 : Let O be the origin and OP and OQ be the tangents to the circle x2+y2-6x+4y+8=0 at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point (α,12), then a value of α is [2023]

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Q 1 :

A square is inscribed in the circle $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$ . One side of this square is parallel to y = x + 3. If $(x_{i}, y_{i})$ are the vertices of the square, then $\sum_{}^{} (x_{i}^{2}, y_{i}^{2})$ is equal to [2024]

Q 2 :

Let C be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line x + y = 2 intersects the circle C at the points P and Q. Let MN be a chord of C of length 2 units and slope –1. Then a distance (in units) between the chord PQ and the chord MN is [2024]

Q 3 :

Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3, 2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5, 5) is: [2024]

Q 4 :

Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at (–1, 0) and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the y-axis at the point P, then the square of the distance of P from the centre of $C_{1}$ is: [2024]

Q 5 :

Let ABCD and AEFG be squares of sides 4 and 2 units respectively. the point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies : [2024]

Q 7 :

If the image of the point (–4,5) in the line x + 2y = 2 lies on the circle ${(x + 4)}^{2} + {(y - 3)}^{2} = r^{2}$ , then r is equal to : [2024]

Q 8 :

Let a circle passing through (2,0) have its centre at the point (h, k). Let $(x_{c}, y_{c})$ be the point of intersection of the lines 3x + 5y = 1 and $(2 + c) x + 5 c^{2} y = 1$ . If $h = \lim_{c \to 1} x_{c}$ and $k = \lim_{c \to 1} y_{c}$ , then the equation of the circle is : [2024]

Q 9 :

Let $C : x^{2} + y^{2} = 4$ and $C' : x^{2} + y^{2} - 4 λ x + 9 = 0$ be two circles. If the set of all values of $λ$ so that the circles C and C' intersect at two distinct points, is R – [a, b], then the point (8a + 12, 16b – 20) lies on the curve : [2024]

Q 10 :

Let the locus of the midpoints of the chords of the circle $x^{2} + {(y - 1)}^{2} = 1$ drawn from the origin intersect the line x + y = 1 at P and Q. Then, the length of PQ is : [2024]

Q 11 :

Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to : [2024]

Q 12 :

If the circles ${(x + 1)}^{2} + {(y + 2)}^{2} = r^{2}$ and $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ intersect at exactly two distinct points, then [2024]

Q 13 :

If one of the diameters of the circle $x^{2} + y^{2} - 10 x + 4 y + 13 = 0$ is a chord of another circle C, whose centre is the point of intersection of the lines 2x + 3y = 12 and 3x –2y = 5, then the radius of the circle C is : [2024]

Q 14 :

Let a variable line passing through the centre of the circle $x^{2} + y^{2} - 16 x - 4 y = 0$ , meet the positive co-ordinate axes at the points A and B. Then the minimum value of OA + OB, where O is the origin is equal to [2024]

Q 15 :

Let the maximum and minimum values of ${(\sqrt{8 x - x^{2} - 12} - 4)}^{2} + {(x - 7)}^{2}, x \in R$ be M and m, respectively. Then $M^{2} - m^{2}$ is equal to __________. [2024]

Q 16 :

Let the centre of a circle, passing through the points (0, 0), (1, 0) and touching the circle $x^{2} + y^{2} = 9$ , be (h, k). Then for all possible values of the coordinates of the centre (h, k), $4 (h^{2} + k^{2})$ is equal to _________. [2024]

Q 17 :

Consider a circle ${(x - α)}^{2} + {(y - β)}^{2} = 50$ , where $α, β > 0$ . If the circle touches the line y + x = 0 at the point P, whose distance from the origin is $4 \sqrt{2}$ , then ${(α + β)}^{2}$ is equal to __________. [2024]

Q 18 :

Equations of two diameters of a circle are 2x – 3y = 5 and 3x – 4y = 7. The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to __________. [2024]

Q 20 :

If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then $10 k + r^{2}$ is equal to [2025]

Q 21 :

Let $C_{1}$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $C_{2}$ be the circle with centre (1, 3) that touches $C_{1}$ externally at the point $(α, β)$ . If ${(β - α)}^{2} = \frac{m}{n}$ , gcd (m, n) = 1, then m + n is equal to [2025]

Q 24 :

Let the equation of the circle, which touches x-axis at the point (a, 0), a > 0 and cuts off an intercept of length b on y-axis be $x^{2} + y^{2} - α x + β y + γ = 0$ . If the circle lies below x-axis, then the ordered pair ( $2 a, b^{2}$ ) is equal to [2025]

Q 25 :

Let the line x + y = 1 meet the circle $x^{2} + y^{2} = 4$ at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to: [2025]

Q 26 :

Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C, whose mid-point is (1, 2), is: [2025]

Q 27 :

The absolute difference between the squares of the radii of the two circles passing through the point (–9, 4) and touching the lines x + y = 3 and x – y = 3, is equal to __________. [2025]

Q 28 :

Let r be the radius of the circle, which touches x-axis at point (a, 0), a < 0 and the parabola $y^{2} = 9 x$ at the point (4, 6). Then r is equal to __________. [2025]

Q 29 :

If the tangents at the point P and Q on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle PQR is [2023]

Q 30 :

Let O be the origin and OP and OQ be the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point $(α, \frac{1}{2}),$ then a value of $α$ is [2023]