JEE Advanced Maths PYQ – Conic Sections (Circles) | Previous Year Questions with Solutions

Q 11 :

Let $M = {(x, y) \in ℝ \times ℝ : x^{2} + y^{2} \leq r^{2}}$ where $r > 0$ .

Consider the geometric progression $a_{n} = \frac{1}{2^{n - 1}}, n = 1, 2, 3, \dots$ . Let $S_{0} = 0$ and, for $n \geq 1$ , let $S_{n}$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$ , let $C_{n}$ denote the circle with center $(S_{n - 1}, 0)$ and radius $a_{n}$ , and $D_{n}$ denote the circle with center $(S_{n - 1}, S_{n - 1})$ and radius $a_{n}$ . [2021]

Q. Consider $M$ with $r = \frac{(2^{199} - 1) \sqrt{2}}{2^{198}}$ . The number of all those circles $D_{n}$ that are inside $M$ is

198
199
200
201

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$∵$ $r = \frac{(2^{199} - 1) \sqrt{2}}{2^{198}}$

$Now, \sqrt{2} S_{n - 1} + a_{n} < \frac{2^{199} - 1}{2^{198}} \sqrt{2}$

$\Rightarrow 2 \sqrt{2} (1 - \frac{1}{2^{n - 1}}) + \frac{1}{2^{n - 1}} < \frac{2^{199} - 1}{2^{198}} \sqrt{2}$

$\Rightarrow \frac{2 \sqrt{2} - 1}{2 \cdot 2^{n - 2}} > \frac{\sqrt{2}}{2^{198}}$

$\Rightarrow 2^{n - 2} < (2 - \frac{1}{\sqrt{2}}) 2^{197} \Rightarrow n \leq 199 \Rightarrow n = 199$

Q 12 :

Let $S$ be the circle in the $x y$ -plane defined by the equation $x^{2} + y^{2} = 4$ . [2018]

Q. Let $E_{1} E_{2}$ and $F_{1} F_{2}$ be the chords of $S$ passing through the point $P_{0} (1, 1)$ and parallel to the $x$ -axis and the $y$ -axis, respectively. Let $G_{1} G_{2}$ be the chord of $S$ passing through $P_{0}$ and having slope $- 1$ . Let the tangents to $S$ at $E_{1}$ and $E_{2}$ meet at $E_{3}$ , the tangents to $S$ at $F_{1}$ and $F_{2}$ meet at $F_{3}$ , and the tangents to $S$ at $G_{1}$ and $G_{2}$ meet at $G_{3}$ . Then, the points $E_{3}, F_{3}$ and $G_{3}$ lie on the curve

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

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$Equation of E_{1} E_{2} is y = 1; Equation of F_{1} F_{2} is x = 1$

$Equation of G_{1} G_{2} is x + y = 2$

$By symmetry, tangents at E_{1} and E_{2} meet on y-axis and tangents at F_{1} and F_{2} meet on x-axis.$

$E_{1} \equiv (\sqrt{3}, 1), F_{1} \equiv (1, \sqrt{3})$

$Equation of tangent at E_{1} is$ $\sqrt{3} x + y = 4$

$Equation of tangent at F_{1} is$ $x + \sqrt{3} y = 4$

$∴$ $Points E_{3} (0, 4) and F_{3} (4, 0)$

$Tangents at G_{1} and G_{2} are x = 2 and y = 2 respectively, intersecting each other at G_{3} (2, 2)$

$Clearly E_{3}, F_{3} and G_{3} lie on the curve x + y = 4 .$

Q 13 :

Let $S$ be the circle in the $xy$ -plane defined by the equation $x^{2} + y^{2} = 4$ . [2018]

Q. Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

${(x + y)}^{2} = 3 x y$
$x^{2 / 3} + y^{2 / 3} = 2^{4 / 3}$
$x^{2} + y^{2} = 2 x y$
$x^{2} + y^{2} = x^{2} y^{2}$

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$Let point P be (2 \cos θ, 2 \sin θ)$

$Tangent at P is$ $x \cos θ + y \sin θ = 2$

$∴$ $M (\frac{2}{\cos θ}, 0) and N (0, \frac{2}{\sin θ})$

$∴$ $Midpoint of M N =$ $(\frac{1}{\cos θ}, \frac{1}{\sin θ})$

$For locus of midpoint (x, y) of M N,$

$x = \frac{1}{\cos θ}, y = \frac{1}{\sin θ}$

$\Rightarrow \frac{1}{x^{2}} + \frac{1}{y^{2}} = 1 \Rightarrow x^{2} + y^{2} = x^{2} y^{2}$

Q 14 :

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A possible equation of $L$ is

$x - \sqrt{3} y = 1$
$x + \sqrt{3} y = 1$
$x - \sqrt{3} y = - 1$
$x + \sqrt{3} y = 5$

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$Equation of tangent P T to the circle x^{2} + y^{2} = 4$ $at the point P (\sqrt{3}, 1) is$ $x \sqrt{3} + y = 4$

$Let the line L, perpendicular to tangent P T be$

$x - \sqrt{3} y + λ = 0$

$As it is tangent to the circle {(x - 3)}^{2} + y^{2} = 1$

$∴$ Length of perpendicular from centre of circle to the tangent = radius of circle

$\Rightarrow$ $| \frac{3 + λ}{2} |$ $= 1 \Rightarrow λ = - 1 or - 5$

$∴$ $Equation of L can be$ $x - \sqrt{3} y = 1 or x - \sqrt{3} y = 5$

Q 15 :

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ , is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A common tangent of the two circles is

$x = 4$
$y = 2$
$x + \sqrt{3} y = 4$
$x + 2 \sqrt{2} y = 6$

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From the figure it is clear that the intersection point of two direct common tangents lies on the $x$ -axis.

$Also ∆ P T_{1} C_{1} ~ ∆ P T_{2} C_{2} \Rightarrow P C_{1} : P C_{2} = 2 : 1$

$Hence, P divides C_{1} C_{2} externally in the ratio 2 : 1$

$∴$ $Coordinates of P = (6, 0)$

$Let the equation of tangent through P be$ $y = m (x - 6)$

$As it touches x^{2} + y^{2} = 4$

$∴$ $| \frac{6 m}{\sqrt{m^{2} + 1}} |$ $= 2 \Rightarrow 36 m^{2} = 4 (m^{2} + 1) \Rightarrow m = \pm \frac{1}{2 \sqrt{2}}$

$∴$ $Equations of common tangents are$

$y = \pm \frac{1}{2 \sqrt{2}} (x - 6)$

$Also x = 2 is a common tangent to the two circles.$

Q 16 :

$A B C D$ is a square of side length 2 units. $C_{1}$ is the circle touching all the sides of the square $A B C D$ and $C_{2}$ is the circumcircle of square $A B C D$ . $L$ is a fixed line in the same plane. [2006]

Q. If $P$ is any point of $C_{1}$ and $Q$ is another point on $C_{2}$ , then $\frac{P A^{2} + P B^{2} + P C^{2} + P D^{2}}{Q A^{2} + Q B^{2} + Q C^{2} + Q D^{2}}$ is equal to

0.75
1.25
1
0.5

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According to the given question, we can assume the square $A B C D$ with its vertices $A (1, 1), B (- 1, 1), C (- 1, - 1), D (1, - 1) .$

$P be the point (0, 1) and Q be the point (\sqrt{2}, 0) .$

Then, $\frac{P A^{2} + P B^{2} + P C^{2} + P D^{2}}{Q A^{2} + Q B^{2} + Q C^{2} + Q D^{2}}$

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

$= \frac{12}{16} = 0.75$

Q 17 :

$A B C D$ is a square of side length 2 units. $C_{1}$ is the circle touching all the sides of the square $A B C D$ and $C_{2}$ is the circumcircle of square $A B C D$ . $L$ is a fixed line in the same plane. [2006]

Q. If a circle is such that it touches the line $L$ and the circle $C_{1}$ externally, such that both the circles are on the same side of the line, then the locus of centre of the circle is

ellipse
hyperbola
parabola
pair of straight line

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Let $C'$ be the said circle touching $C_{1}$ and $L$ , so that $C_{1}$ and $C'$ are on the same side of $L$ . Let us draw a line $T$ parallel to $L$ at a distance equal to the radius of circle $C_{1},$ on the opposite side of $L$ .

Then the centre of $C'$ is equidistant from the centre of $C_{1}$ and from line $T$ .

$\Rightarrow$ Locus of centre of $C'$ is a parabola.

Q 18 :

$A B C D$ is a square of side length 2 units. $C_{1}$ is the circle touching all the sides of the square $A B C D$ and $C_{2}$ is the circumcircle of square $A B C D$ . $L$ is a fixed line in the same plane. [2006]

Q. A line $L'$ through $A$ is drawn parallel to $B D$ . Point $S$ moves such that its distances from the line $B D$ and the vertex $A$ are equal. If locus of $S$ cuts $L'$ at $T_{2}$ and $T_{3}$ and $A C$ at $T_{1}$ , then area of $∆ T_{1} T_{2} T_{3}$ is

$\frac{1}{2}$ sq. units
$\frac{2}{3}$ sq. units
$1$ sq. units
$2$ sq. units

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Since $S$ is equidistant from $A$ and line $B D$ , it traces a parabola.

Clearly, $A C$ is the axis, $A (1, 1)$ is the focus and $T_{1} (\frac{1}{2}, \frac{1}{2})$ is the vertex of the parabola.

$A T_{1} = \frac{1}{\sqrt{2}}$

$T_{2} T_{3} = latus rectum of parabola = 4 \times \frac{1}{\sqrt{2}} = 2 \sqrt{2}$

$∴$ $Area (∆ T_{1} T_{2} T_{3}) = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times 2 \sqrt{2} = 1 sq. units$

Q 19 :

Consider $L_{1} : 2 x + 3 y + p - 3 = 0$

$L_{2} : 2 x + 3 y + p + 3 = 0$

where $p$ is a real number, and $C : x^{2} + y^{2} + 6 x - 10 y + 30 = 0$

STATEMENT - 1: If line $L_{1}$ is a chord of circle $C$ , then line $L_{2}$ is not always a diameter of circle $C$ .

STATEMENT - 2: If line $L_{1}$ is a diameter of circle $C$ , then line $L_{2}$ is not a chord of circle $C$ . [2008]

Statement - 1 is True, Statement - 2 is True; Statement - 2 is a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is False
Statement - 1 is False, Statement - 2 is True

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Given: A circle $x^{2} + y^{2} + 6 x - 10 y + 30 = 0$ with centre $(- 3, 5)$ and radius = 2

$L_{1} : 2 x + 3 y + (p - 3) = 0$

$L_{2} : 2 x + 3 y + p + 3 = 0$

$Clearly L_{1} ∥ L_{2}$

$∴$ $Distance between L_{1} and L_{2}$

$=$ $| \frac{(p + 3) - (p - 3)}{\sqrt{2^{2} + 3^{2}}} |$ $= \frac{6}{\sqrt{13}} < 2$

$\Rightarrow$ If one line is a chord of the given circle, the other line may or may not be the diameter of the circle.

$∴$ Statement 1 is true and Statement 2 is false.

Q 20 :

Tangents are drawn from the point (17, 7) to the circle $x^{2} + y^{2} = 169$ .

STATEMENT-1: The tangents are mutually perpendicular.

because

STATEMENT-2: The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is $x^{2} + y^{2} = 338$ . [2007]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is False.
Statement-1 is False, Statement-2 is True.

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Equation of director circle of the given circle $x^{2} + y^{2} = 169$ is $x^{2} + y^{2} = 2 \times 169 = 338$

We know that from every point on the director circle, the tangents drawn to the given circle are perpendicular to each other.

Since (17,7) lies on the director circle,

$∴$ The tangents from (17,7) to the given circle are mutually perpendicular.

Topic Question Set

Q 14 :

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A possible equation of $L$ is

Q 15 :

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ , is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A common tangent of the two circles is

Q 20 :

Tangents are drawn from the point (17, 7) to the circle $x^{2} + y^{2} = 169$ .

STATEMENT-1: The tangents are mutually perpendicular.

because

STATEMENT-2: The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is $x^{2} + y^{2} = 338$ . [2007]

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Topic Question Set

Q 14 : A tangent PT is drawn to the circle x2+y2=4 at the point P(3,1). A straight line L, perpendicular to PT is a tangent to the circle (x-3)2+y2=1. [2012] Q. A possible equation of L is

Q 15 : A tangent PT is drawn to the circle x2+y2=4 at the point P(3,1). A straight line L, perpendicular to PT, is a tangent to the circle (x-3)2+y2=1. [2012] Q. A common tangent of the two circles is

Q 16 : ABCD is a square of side length 2 units. C1 is the circle touching all the sides of the square ABCD and C2 is the circumcircle of square ABCD. L is a fixed line in the same plane. [2006] Q. If P is any point of C1 and Q is another point on C2, then PA2+PB2+PC2+PD2QA2+QB2+QC2+QD2 is equal to

Q 19 : Consider L1:2x+3y+p-3=0 L2:2x+3y+p+3=0 where p is a real number, and C:x2+y2+6x-10y+30=0 STATEMENT - 1: If line L1 is a chord of circle C, then line L2 is not always a diameter of circle C. STATEMENT - 2: If line L1 is a diameter of circle C, then line L2 is not a chord of circle C. [2008]

Q 20 : Tangents are drawn from the point (17, 7) to the circle x2+y2=169. STATEMENT-1: The tangents are mutually perpendicular. because STATEMENT-2: The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2+y2=338. [2007]

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Q 14 :

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A possible equation of $L$ is

Q 15 :

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ , is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A common tangent of the two circles is

Q 20 :

Tangents are drawn from the point (17, 7) to the circle $x^{2} + y^{2} = 169$ .

STATEMENT-1: The tangents are mutually perpendicular.

because

STATEMENT-2: The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is $x^{2} + y^{2} = 338$ . [2007]