Circle – JEE Main Previous Year Question with Solutions
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Q. Three distinct points A, B and C are given in the 2βdimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (β1, 0) is equal to $\frac{1}{3}$. Then the circumcentre of the triangle ABC is at the point :- (1) $\left(\frac{5}{2}, 0\right)$ ( 2)$\left(\frac{5}{3}, 0\right)$ (3) (0, 0) ( 4)$\left(\frac{5}{4}, 0\right)$ [AIEEE-2009]

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Sol. (4) Let $A, B, C$ are represented by the point $(x, y)$ $\frac{\sqrt{(x-1)^{2}+y^{2}}}{\sqrt{(x+1)^{2}+y^{2}}}=\frac{1}{2}$ $8 x^{2}+8 y^{2}-20 x+8=0$ Which is the circle which passes through the points $A, B, C$ then circumcentre will be the centre of the circle $\left(\frac{5}{4}, 0\right)$

Q. If $P$ and $Q$ are the points of intersection of the circles $x^{2}+y^{2}+3 x+7 y+2 p-5=0$ and $x^{2}+y^{2}+2 x+2 y-p^{2}=0,$ then there is a circle passing through $P, Q$ and $(1,1)$ is possible for : $-$ (1) All except two values of p (2) Exactly one value of p (3) All values of p (4) All except one value of p [AIEEE-2009]

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Sol. (4) Equation of line $\mathrm{PQ}$ $\mathrm{x}+5 \mathrm{y}+2 \mathrm{p}-5+\mathrm{p}^{2}=0$ $\mathrm{P}, \mathrm{Q}$ and $(1,1)$ will not lie on a circle of $(1,1)$ Lies on the line $\mathrm{x}+5 \mathrm{y}+\mathrm{p}^{2}+2 \mathrm{p}-5=0$

Q. For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is :- (1) There is a regular polygon with $\frac{\mathrm{r}}{\mathrm{R}}=\frac{1}{2}$ (2) There is a regular polygon with $\frac{r}{R}=\frac{1}{\sqrt{2}}$ (3) There is a regular polygon with $\frac{\mathrm{r}}{\mathrm{R}}=\frac{2}{3}$b (4) There is a regular polygon with $\frac{\mathrm{r}}{\mathrm{R}}=\frac{\sqrt{3}}{2}$ [AIEEE-2010]

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Sol. (3)

Q. The circle $\mathrm{x}^{2}+\mathrm{y}^{2}=4 \mathrm{x}+8 \mathrm{y}+5$ intersects the line $3 \mathrm{x}-4 \mathrm{y}=\mathrm{m}$ at two distinct points if :- (1) β 85 < m < β 35 (2) β 35 < m < 15 (3) 15 < m < 65 (4) 35 < m < 85 [AIEEE-2010]

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Sol. (2)

Q. The two circles $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{ax}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}(\mathrm{c}>0)$ touch each other if :- (1) $\mathrm{a}=2 \mathrm{c}$ (2) $|\mathrm{a}|=2 \mathrm{c}$ (3) $2|\mathrm{a}|=\mathrm{c}$ (4) $|\mathrm{a}|=\mathrm{c}$ [AIEEE-2011]

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Sol. (4)

Q. The equation of the circle passing through the points (1, 0) and (0, 1) and having the smallest radius is β (1) $x^{2}+y^{2}+x+y-2=0$ (2) $x^{2}+y^{2}-2 x-2 y+1=0$ (3) $x^{2}+y^{2}-x-y=0$ (4) $x^{2}+y^{2}+2 x+2 y-7=0$ [AIEEE-2011]

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Sol. (3) (1, 0) and (0, 1) will be ends of diameter So equation of circle (x- 1) ( $\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-1)$ $\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{x}-\mathrm{y}=0$

Q. The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is : (1) 5/3Β  Β  Β  Β  Β  Β (2) 10/3Β  Β  Β  Β  Β  Β  (3) 3/5Β  Β  Β  Β  Β  Β  Β (4) 6/5 [AIEEE-2012]

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Sol. (2) diameter $=\frac{10}{3}$

Q. The circle passing through (1, β 2) and touching the axis of x at (3, 0) also passes through the pointΒ : (1) (β5, 2)Β  Β  Β  Β  Β  (2) (2, β5)Β  Β  Β  Β  (3) (5, β2)Β  Β  Β  Β  Β  (4) (β2, 5) [JEE (Main)-2013]

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Sol. (3) Let equation of circle be $(\mathrm{x}-3)^{2}+(\mathrm{y}+\mathrm{r})^{2}=\mathrm{r}^{2} \because$ it passes through $(1,-2)$ $\Rightarrow \mathrm{r}=2$ $\Rightarrow \quad$ circle is $(\mathrm{x}-3)^{2}+(\mathrm{y}+2)^{2}=4$ $\Rightarrow \quad(5,-2)$ Aliter : $(\mathrm{x}-3)^{2}+\mathrm{y}^{2}+\lambda \mathrm{y}=0 \quad \ldots .(1)$ Putting $(1,-2)$ in ( 1) $\Rightarrow \lambda=4$ Required circle is $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}+4 \mathrm{y}+9=0$ point $(5,-2)$ satisfies the equation the equation

Q. If a circle $C$ passing through $(4,0)$ touches the circle $x^{2}+y^{2}+4 x-6 y-12=0$ externally at a point $(1,-1),$ then the radius of the circle $C$ is :- (1) $\sqrt{57}$ (2) $2 \sqrt{5}$ (3) 4 (4) 5 [JEE (Main)-2013]

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Sol. (4) Equation of family of circles is $(\mathrm{x}-1)^{2}+(\mathrm{y}+1)^{2}+\lambda\left(\mathrm{x}^{2}+\mathrm{y}^{2}+4 \mathrm{x}-6 \mathrm{y}-12\right)=0$ Now it passes of through $(4,0)$ we get $\Rightarrow 9+1+\lambda(16+16-12)=0 \Rightarrow \lambda=-\frac{10}{20}=-\frac{1}{2}$ Required equation of circle is $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+2 \mathrm{y}+2-\frac{1}{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+4 \mathrm{x}-6 \mathrm{y}-12\right)=0$ $\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-8 \mathrm{x}+10 \mathrm{y}+16=0$ radius $=\sqrt{16+25-16}=5$

Q. If the circle $x^{2}+y^{2}-6 x-8 y+\left(25-a^{2}\right)=0$ touches the axis of $x,$ then a equals :- $(1) \pm 4$ $(2) \pm 3$ (3) 0 $(4) \pm 2$ [JEE-Main (on line)-2013]

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Sol. (3,4)

Q. Statement I : The only circle having radius $\sqrt{10}$ and a diameter along line $2 \mathrm{x}+\mathrm{y}=5$ is $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}+2 \mathrm{y}=0$ Statement $\mathbf{I I}: 2 \mathbf{x}+\mathrm{y}=5$ is a normal to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}+2 \mathrm{y}=0$. [JEE-Main (on line)-2013] (1) Statement I is false, Statement II is true (2) Statement I is true ; Statement II is false. (3) Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I. (4) Statement I is true : Statement II is true ; Statement II is a correct explanation for Statement I. [JEE-Main (on line)-2013]

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Sol. (1) Normal passes through centre of circle but there can be more then one circle.

Q. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal toΒ : (1) $\frac{\sqrt{3}}{\sqrt{2}}$ ( 2)$\frac{\sqrt{3}}{2}$ (3) $\frac{1}{2}$ (4) $\frac{1}{4}$ [JEE(Main)-2014]

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Sol. (4) $\mathrm{C}_{1} \mathrm{C}_{2}=\mathrm{r}_{1}+\mathrm{r}_{2}$ $\sqrt{1+(1-\mathrm{y})^{2}}=1+\mathrm{y}$ $\therefore \mathrm{y}=\frac{1}{4}$ $\therefore \mathrm{radius}=\frac{1}{4}$

Q. The number of common tangents to the circle $x^{2}+y^{2}-4 x-6 y-12=0$ and $x^{2}+y^{2}+6 x$ $+18 y+26=0,$ is : (1) 3Β  Β  Β  Β  Β  Β  Β (2) 4Β  Β  Β  Β  Β  Β  Β  Β (3) 1Β  Β  Β  Β  Β  Β  Β  Β (4) 2 [JEE(Main)-2015]

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Sol. (1) distance between centres $\left(\mathrm{c}_{1} \mathrm{c}_{2}\right)=13$ $\mathrm{r}_{1}=\sqrt{4+9+12}=5 \& \mathrm{r}_{2}=\sqrt{9+81-26}=8$ $\mathrm{c}_{1} \mathrm{c}_{2}=\mathrm{r}_{1}+\mathrm{r}_{2}$ is a case of external touching (as shown in above figure), so three common tangentscan be drawn.

Q. If one of the diameters of the circle, given by the euqation, $x^{2}+y^{2}-4 x+6 y-12=0,$ is a chord of a circle $S,$ whose centre is at $(-3,2),$ then the radius of $S$ is : $-$ (1) 10 (2) $5 \sqrt{2}$ (3) $5 \sqrt{3}$ (4) 5 [JEE(Main)-2016]

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Sol. (3)

Q. The centres of those circles which touch the circle, $x^{2}+y^{2}-8 x-8 y-4=0,$ externally and also touch the $x$ -axis, lie on :- (1) A parabola (2) A circle (3) An ellipse which is not a circle (4) A hyperbola [JEE(Main)-2016]

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Sol. (1)

Q. If the tangent at $(1,7)$ to the curve $x^{2}=y-6$ touches the circle $x^{2}+y^{2}+16 x+12 y+c=0$ then the value of $c$ is : (1) 185Β  Β  Β  Β  Β (2) 85Β  Β  Β  Β  Β  Β  (3) 95Β  Β  Β  Β  Β  Β  (4) 195 [JEE(Main)-2018]

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Sol. (3)

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120
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1
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4
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0
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0
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0
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