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

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

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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]

0%

(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]

0%

(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

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

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

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

Topic Question Set

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]

0%

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]

0%

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Topic Question Set

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]

0%

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]

0%

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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]