Circle JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Circle, free to read — no sign-in needed. The full chapter has 30 questions; sign in to attempt the remaining 26 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Advanced Adv 2014 (Paper 1)
   A circle S passes through the point (0, 1) and is orthogonal to the circles ${\left(x-1\right)}^{2}+{y}^{2}=16$ and ${x⁡}^{2}+{y⁡}^{2}=1$. Then
   1. A.Radius of S is 8
   2. B.Radius of S is 7
   3. C.Centre of S is (-7, 1)
   4. D.Centre of S is (-8, 1)
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   Answer: B,C

   Given circles ${x}^{2}+{y}^{2}-2x-15=0$ ${x}^{2}+{y}^{2}-1=0$ Radical axis $x+7=0$ ......(i) Centre of circle lies on (i) Let the centre be $\left(-7,k\right)$ Let equation be ${x}^{2}+{y}^{2}+14x-2ky+c=0$ Orthogonallity gives $-14=c-15\Rightarrow c=1$ .......(ii) $\left(0,1\right)\rightarrow 1-2k+1=0\Rightarrow k=1$ Hence radius $=\sqrt{{7}^{2}+{k}^{2}-c}=\sqrt{49+1-1}=7$ Alternate Solution Given circles ${x}^{2}+{y}^{2}-2x-15=0$ ${x}^{2}+{y}^{2}-1=0$ Let equation of circle ${x}^{2}+{y}^{2}+2gx+2fy+c=0$ Circle passes through (0, 1) $\Rightarrow$ $1+2f+c=0$ Applying condition of orthogonality $-2g=c-15,0=c-1$ $\Rightarrow$ $c=1,g=7,f=-1$ $r=\sqrt{49+1-1}=7;$ centre $\left(-7,1\right)$
2. Q2JEE Advanced Adv 2012 (Paper 2)
   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 $(x-3)^{2}+y^{2}-1$. Question: A possible equation of $L$ is
   1. A.$x-\sqrt{3} y=1$
   2. B.$x+\sqrt{3} y=1$
   3. C.$x-\sqrt{3} y=-1$
   4. D.$x+\sqrt{3} y=5$
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   Answer: (A)

   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 $x-y \sqrt{3}+\lambda=0$ As it is tangent to the circle $(x-3)^{2}+y^{2}=1$ $\therefore \quad$ Length of perpendicular from centre of circle to the Tangent $=$ radius of circle. $\Rightarrow\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1 \text { or }-5$ $\therefore$ Equation of $L$ can be $x-\sqrt{3} y=1$ or $x-\sqrt{3} y=5$
3. Q3JEE Advanced Adv 2011 (Paper 2)
   The circle passing through the point $(-1,0)$ and touching the $Y$-axis at $(0,2)$, also passes through the point
   1. A.$\left(-\frac{3}{2}, 0\right)$
   2. B.$\left(-\frac{5}{2}, 2\right)$
   3. C.$\left(-\frac{3}{2}, \frac{5}{2}\right)$
   4. D.$(-1,-4)$
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   Answer: (D)

   Equation of circle passing through a point $\left(x_1, y_1\right)$ and touching the straight line $L$, is given by $$\left(x-x_1\right)^2+\left(y-y_1\right)^2=\lambda L=0$$ $\therefore$ Equation of circle passing through $(0,2)$ and touching $x=0$. Now, $(x-0)^2+(y-2)^2+\lambda x=0 \ldots$ (i) Also, it passes through $(-1,0)$. So, $1+4-\lambda=0 \Rightarrow \lambda=5$ Eq. (i) becomes, $$\begin{aligned} & x^2+y^2-4 y+4+5 x=0 \\ \Rightarrow \quad x^2+y^2+5 x-4 y+4 & =0 \end{aligned}$$ For $x$-intercept, put $y=0$, $$\begin{array}{rlrl} & x^2+5 x+4 & =0 \\ \Rightarrow & & (x+1)(x+4) & =0 \\ \therefore & & x & =-1,-4 \end{array}$$
4. Q4JEE Advanced Adv 2010 (Paper 1)
   Paragraph: The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question: Equation of the circle with $A B$ as its diameter is
   1. A.$x^2+y^2-12 x+24=0$
   2. B.$x^2+y^2+12 x+24=0$
   3. C.$x^2+y^2+24 x-12=0$
   4. D.$x^2+y^2-24 x-12=0$
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   Answer: (A)

   The equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$ and that of circle is $$x^2+y^2-8 x=0$$ For their points of intersection $$\frac{x^2}{9}+\frac{x^2-8 x}{4}=1$$ $$\begin{array}{ll} \Rightarrow & 4 x^2+9 x^2-72 x=36 \\ \Rightarrow & 13 x^2-72 x-36=0 \end{array}$$ $$\begin{aligned} & \Rightarrow \quad 13 x^2-78 x+6 x-36=0 \\ & \Rightarrow \quad 13 x(x-6)+6(x-6)=0 \\ & \Rightarrow \quad x=6, x=-\frac{13}{6} \\ & x=-\frac{13}{6} \text { not acceptable } \\ & \end{aligned}$$ Now, for $x=6, y=\pm 2 \sqrt{3}$ Required equation is $$\begin{aligned} & (x-6)^2+(y+2 \sqrt{3})(y-2 \sqrt{3})=0 \\ & \Rightarrow \quad x^2-12 x+y^2+24=0 \\ & \Rightarrow \quad x^2+y^2-12 x+24=0 \end{aligned}$$