Circle – JEE Advanced Previous Year Questions with Solutions

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Q. Tangents drawn from the point $\mathrm{P}(1,8)$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}-4 \mathrm{y}-11=0$ touch the circle at the points $\mathrm{A}$ and $\mathrm{B}$. The equation of the circumcircle of the triangle $\mathrm{PAB}$ is

(A) $\mathrm{x}^{2}+\mathrm{y}^{2}+4 \mathrm{x}-6 \mathrm{y}+19=0$

(B) $\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-10 \mathrm{y}+19=0$

(C) $x^{2}+y^{2}-2 x+6 y-29=0$

(D) $x^{2}+y^{2}-6 x-4 y+19=0$

[JEE 2009, 3]

Sol. (B)

Q. The centres of two circles $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ each of unit radius are at a distance of 6 units from eachother. Let $\mathrm{P}$ be the mid point of the line segment joining the centres of $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ and $\mathrm{C}$ be a circle touching circles $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ externally. If a common tangent to $\mathrm{C}_{1}$ and $\mathrm{C}$ passing through $\mathrm{P}$ is also a common tangent to $\mathrm{C}_{2}$ and $\mathrm{C}$, then the radius of the circle $\mathrm{C}$ is

[JEE 2009, 4]

Sol. 8

Q. Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the center, angles of $\frac{\pi}{\mathrm{k}}$ and $\frac{2 \pi}{\mathrm{k}},$ where $\mathrm{k}>0,$ then the value of $[\mathrm{k}]$ is

[Note : [k] denotes the largest integer less than or equal to k]

[JEE 10, 3M]

Sol. 3

Q. The circle passing through the point (–1,0) and touching the y-axis at (0,2) also passes through the point –

(A) $\left(-\frac{3}{2}, 0\right)$

(B) $\left(-\frac{5}{2}, 2\right)$

(C) $\left(-\frac{3}{2}, \frac{5}{2}\right)$

(D) (–4,0)

[JEE 2011, 3M, –1M]

Sol. (D)

Family of circle which touches $y$ -axis at $(0,2)$ is

$x^{2}+(y-2)^{2}+\lambda x=0$

Passing through $(-1,0)$

$\Rightarrow 1+4-\lambda=0 \quad \Rightarrow \quad \lambda=5$

$\therefore \quad x^{2}+y^{2}+5 x-4 y+4=0$

$\quad$ which satisfy the point $(-4,0)$

Q. The straight line $2 \mathrm{x}-3 \mathrm{y}=1$ divides the circular region $\mathrm{x}^{2}+\mathrm{y}^{2} \leq 6$ into two parts. If $S=\left\{\left(2, \frac{3}{4}\right),\left(\frac{5}{2}, \frac{3}{4}\right),\left(\frac{1}{4},-\frac{1}{4}\right),\left(\frac{1}{8}, \frac{1}{4}\right)\right\},$ then the number of point(s) in Slying inside the smaller part is

[JEE 2011, 4M]

Sol. 2

for origin : 2 × 0 – 3 × 0 – 1 = – 1 (–ve)

Q. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4 \mathrm{x}-5 \mathrm{y}=20$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=9 \mathrm{is}-$

(A) $20\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)-36 \mathrm{x}+45 \mathrm{y}=0$

(B) $20\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+36 \mathrm{x}-45 \mathrm{y}=0$

(C) $36\left(x^{2}+y^{2}\right)-20 x+45 y=0$

(D) $36\left(x^{2}+y^{2}\right)+20 x-45 y=0$

[JEE 2012, 3M, –1M]

Sol. (A)

Paragraph for Question 7 and 8

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

Q. A common tangent of the two circles is

(A) x = 4

(B) y = 2

(C) $x+\sqrt{3} y=4$

(D) $x+2 \sqrt{2} y=6$

[JEE 2012, 3M, –1M]

Sol. (D)

Q. A possible equation of L is

(A) $x-\sqrt{3} y=1$

(B) $x+\sqrt{3} y=1$

(C) $x-\sqrt{3} y=-1$

(D) $x+\sqrt{3} y=5$

[JEE 2012, 3M, –1M]

Sol. (A)

Equation of tangent at $P$ will be $\sqrt{3} x+y=4$

Slope of line L will be $\frac{1}{\sqrt{3}}$

Let equation of $\mathrm{L}$ be $: y=\frac{\mathrm{x}}{\sqrt{3}}+\mathrm{c}$

$\Rightarrow \quad x-\sqrt{3} y+\sqrt{3} c=0$

Now this $L$ is tangent to $2^{\text {nd }}$ circle

So $\frac{3+\sqrt{3} c}{2}=\pm 1 \quad \Rightarrow \quad c=-\frac{1}{\sqrt{3}}$

or $c=-\frac{5}{\sqrt{3}}$

using $\quad c=-\frac{1}{\sqrt{3}}$

$y=\frac{x}{\sqrt{3}}-\frac{1}{\sqrt{3}} \Rightarrow x-\sqrt{3} y=1 .$ Hence $(\mathrm{A})$

Q. Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length $2 \sqrt{7}$ or y-axis is (are)

(A) $x^{2}+y^{2}-6 x+8 y+9=0$

(B) $x^{2}+y^{2}-6 x+7 y+9=0$

(C) $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}-8 \mathrm{y}+9=0$

(D) $x^{2}+y^{2}-6 x-7 y+9=0$

[JEE(Advanced) 2013, 3, (–1)]

Sol. (A,C)

Q. A circle S passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^{2}+y^{2}=16$ and $x^{2}+y^{2}=1 .$ Then $:-$

(1) radius of S is 8

(B) radius of S is 7

(3) centre of S is (–7, 1)

(D) centre is S is (–8, 1)

[JEE(Advanced)-2014, 3]

Sol. (B)

Q. Let RS be the diameter of the circle x^{2}+y^{2}=1, where S is the point (1,0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. then the locus of E passes through the point(s)-

(A) $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$

(B) $\left(\frac{1}{4}, \frac{1}{2}\right)$

(C) $\left(\frac{1}{3},-\frac{1}{\sqrt{3}}\right)$

(D) $\left(\frac{1}{4},-\frac{1}{2}\right)$ (D) $\left(\frac{1}{4},-\frac{1}{2}\right)$

[JEE(Advanced)-2016]

Sol. (A,C)

Q. For how many values of p, the circle $x^{2}+y^{2}+2 x+4 y-p=0$ and the coordinate axes have exactly three common points ?

[JEE(Advanced)-2017]

Sol. (2)

Q. Let T be the line passing through the points P(–2, 7) and Q(2, –5). Let $\mathrm{F}_{2}$ be the set of all pairs of circles $\left(\mathrm{S}_{1}, \mathrm{S}_{2}\right)$ such that T is tangents to $\mathrm{S}_{1}$ at P and tangent to $\mathrm{S}_{2}$ at Q, and also such that $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ touch each other at a point, say, M. Let $\mathrm{E}_{1}$ be the set representing the locus of M as the pair $\left(\mathrm{S}_{1}, \mathrm{S}_{2}\right)$ varies in $\mathrm{F}_{1}$. Let the set of all straight line segments joining a pair of distinct points of $E_{1}$ and passing through the point R(1, 1) be $\mathrm{F}_{2}$. Let $E_{2}$ be the set of the mid-points of the line segments in the set $\mathrm{F}_{2}$. Then, which of the following statement(s) is (are) TRUE ?

(A) The point $(-2,7)$ lies in $\mathrm{E}_{1}$

(B) The point $\left(\frac{4}{5}, \frac{7}{5}\right)$ does NOT lie in $\mathrm{E}_{2}$

(C) The point $\left(\frac{1}{2}, 1\right)$ lies in $\mathrm{E}_{2}$

(D) The point $\left(0, \frac{3}{2}\right)$ does NOT lie in $\mathrm{E}_{1}$

[JEE(Advanced)-2018]

Sol. (D)

Paragraph “X”

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

(There are two question based on Paragraph “X”, the question given below is one of them)

Q. Let $\mathrm{E}_{1} \mathrm{E}_{2}$ and $\mathrm{F}_{1} \mathrm{F}_{2}$ be the chord of S passing through the point $\mathrm{P}_{0}(1,1)$ and parallel to the $\mathrm{x}-$ axis and the $\mathrm{y}$ -axis, respectively. Let $\mathrm{G}_{1} \mathrm{G}_{2}$ be the chord of S passing through $\mathrm{P}_{0}$ and having slop $-1 .$ Let the tangents to $\mathrm{S}$ at $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ meet at $\mathrm{E}_{3}$, the tangents of $\mathrm{S}$ at $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ meet at $\mathrm{F}_{3},$ and the tangents to $\mathrm{S}$ at $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ meet at $\mathrm{G}_{3} .$ Then, the points $\mathrm{E}_{3}, \mathrm{F}_{3}$ and $\mathrm{G}_{3}$ lie on the curve

(A) x + y = 4

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

(C) (x – 4) (y – 4) = 4

(D) xy = 4

[JEE(Advanced)-2018]

Sol. (A )

Paragraph “X”

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

(There are two questions based on Paragraph “X”, the question given below is one of them)

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 –

(A) $(\mathrm{x}+\mathrm{y})^{2}=3 \mathrm{xy}$

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

(C) $x^{2}+y^{2}=2 x y$

(D) $x^{2}+y^{2}=x^{2} y^{2}$

[JEE(Advanced)-2018]

Sol. 15

Tangent at $\mathrm{P}(2 \cos \theta, 2 \sin \theta)$ is $\mathrm{xcos} \theta+\mathrm{y} \sin \theta=2$

$\mathrm{M}(2 \sec \theta, 0)$ and $\mathrm{N}(0,2 \csc \theta)$

Let midpoint be $(\mathrm{h}, \mathrm{k})$

$\mathrm{h}=\sec \theta, \quad \mathrm{k}=\csc \theta$

$\frac{1}{\mathrm{h}^{2}}+\frac{1}{\mathrm{k}^{2}}=1$

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

Administrator

Comments
• May 27, 2020 at 7:37 pm

Well done

• May 26, 2020 at 4:22 pm

okay good