Circle JEE Main previous year questions with solutions

4 solved JEE Main questions on Circle, free to read — no sign-in needed. The full chapter has 164 questions; sign in to attempt the remaining 160 in the exam simulator.

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1. Q1JEE Main 2026 (28 Jan, Shift 1)Chord with given Middle Point
   Let $y=x$ be the equation of a chord of the circle $\mathrm{C}_{1}$ (in the closed half-plane $x \geq 0$) of diameter 10 passing through the origin. Let $\mathrm{C}_{2}$ be another circle described on the given chord as its diameter. If the equation of the chord of the circle $\mathrm{C}_{2}$, which passes through the point $(2,3)$ and is farthest from the center of $\mathrm{C}_{2}$, is $x+a y+b=0$, then $a-b$ is equal to
   1. A.-2
   2. B.10
   3. C.-6
   4. D.6
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   Answer: (A)

   Circle $C_1$ has radius 5 and chord on line $y = x$ through origin. Setting center at $(5, 0)$ (in region $x \geq 0$), the chord endpoints are $(0, 0)$ and $(5, 5)$ by solving $(t-5)^2 + t^2 = 25$. Circle $C_2$ has this chord as diameter, so center is at $(\frac{5}{2}, \frac{5}{2})$ with radius $\frac{5\sqrt{2}}{2}$. The chord of $C_2$ through $(2, 3)$ that is farthest from center is perpendicular to the radial direction from center to $(2, 3)$. The radial direction is $(2 - \frac{5}{2}, 3 - \frac{5}{2}) = (-\frac{1}{2}, \frac{1}{2})$ with normal direction $(1, -1)$. The chord equation is $1(x - 2) - 1(y - 3) = 0$, giving $x - y + 1 = 0$. In form $x + ay + b = 0$, we have $a = -1$ and $b = 1$, so $a - b = -1 - 1 = -2$.
2. Q2JEE Main 2025 (23 Jan, Shift 2)standard equation of parabola
   <p>The focus of the parabola $y^2=4 x+16$ is the centre of the circle $C$ of radius 5 . If the values of $\lambda$, for which C passes through the point of intersection of the lines $3 x-y=0$ and $x+\lambda y=4$, are $\lambda_1$ and $\lambda_2, \lambda_1 \lt \lambda_2$, then $12 \lambda_1+29 \lambda_2$ is equal to $\qquad$</p>
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   Answer: 15

   <p>$y^2=4(x+4)$ Equation of circle $(x+3)^2+y^2=25$ Passes through the point of intersection of two lines $3 x-y=0$ and $x+\lambda y=4$ $\left(\frac{4}{3 \lambda+1}, \frac{12}{3 \lambda+1}\right) \text {, we get }$ $\begin{aligned} &amp; \lambda=-\frac{7}{6}, 1 \\ &amp; 12 \lambda_1+29 \lambda_2 \\ &amp; -14+29=15 \end{aligned}$</p>
3. Q3JEE Main 2024 (05 Apr, Shift 2)Common Chord
   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 $\mathrm{C}_1$ and $\mathrm{C}_2$ intersects the $y$-axis at the point $\mathrm{P}$, then the square of the distance of $\mathrm{P}$ from the centre of $\mathrm{C}_1$ is :
   1. A.2
   2. B.1
   3. C.4
   4. D.6
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   Answer: (A)

   $\begin{aligned} & S_1: x^2+y^2-2 x-2 y+1=0 \\ & S_2: x^2+y^2+2 x-3=0 \\ & \text { Common chord }=S_1-S_2=0 \\ & -4 x-2 y+4=0 \\ & 2 x+y=2 \Rightarrow P(0,2) \\ & d_{(c, p)}^2=(1-0)^2+(2-1)^2=2\end{aligned}$
4. Q4JEE Main 2023 (15 Apr, Shift 1)Common Tangents
   The number of common tangents, to the circles ${x}^{2}+{y}^{2}-18x-15y+131=0$ and ${x}^{2}+{y}^{2}-6x-6y-7=0$, is
   1. A.$3$
   2. B.$1$
   3. C.$4$
   4. D.$2$
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   Answer: (A)

   If $S\equiv {x}^{2}+{y}^{2}+2gx+2fy+c=0$. $C\equiv \left(-g,-f\right)$ and $r=\sqrt{{g}^{2}+{f}^{2}-c}$. The given circles are ${x}^{2}+{y}^{2}-18x-15y+131=0$ $\Rightarrow {C}_{1}\left(9,\dfrac{15}{2}\right),{r}_{1}=\sqrt{81+\dfrac{225}{4}-131}=\dfrac{5}{2}$ and ${x}^{2}+{y}^{2}-6x-6y-7=0$ $\Rightarrow {C}_{2}\left(3,3\right),{r}_{2}=\sqrt{9+9+7}=5$ $\Rightarrow d={C}_{1}{C}_{2}=\sqrt{{\left(9-3\right)}^{2}+{\left(\dfrac{15}{2}-3\right)}^{2}}=\sqrt{36+\dfrac{81}{4}}=\dfrac{15}{2}$ $\Rightarrow {r}_{1}+{r}_{2}=\dfrac{5}{2}+5=\dfrac{15}{2}$ $\Rightarrow {C}_{1}{C}_{2}={r}_{1}+{r}_{2}$ $∴$ Circles touch each other externally, $3$ common tangents. Hence this is the correct option.