Circle – JEE Main Previous Year Question with Solutions

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Q. Three distinct points A, B and C are given in the 2–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 (–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]

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]

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]

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) – 85 < m < – 35 (2) – 35 < m < 15 (3) 15 < m < 65 (4) 35 < m < 85

[AIEEE-2010]

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]

Sol. (4)

Q. The equation of the circle passing through the points (1, 0) and (0, 1) and having the smallest radius is –

(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]

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) 5/3           (2) 10/3            (3) 3/5             (4) 6/5

[AIEEE-2012]

Sol. (2)

diameter $=\frac{10}{3}$

Q. The circle passing through (1, – 2) and touching the axis of x at (3, 0) also passes through the point :

(1) (–5, 2)          (2) (2, –5)        (3) (5, –2)          (4) (–2, 5)

[JEE (Main)-2013]

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]

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]

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]

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 to :

(1) $\frac{\sqrt{3}}{\sqrt{2}}$

( 2)$\frac{\sqrt{3}}{2}$

(3) $\frac{1}{2}$

(4) $\frac{1}{4}$

[JEE(Main)-2014]

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) 3             (2) 4               (3) 1               (4) 2

[JEE(Main)-2015]

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]

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]

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) 185         (2) 85            (3) 95            (4) 195

[JEE(Main)-2018]

Sol. (3)

• July 1, 2020 at 11:07 am

𝑈𝑝𝑑𝑎𝑡𝑒 𝐽𝐸𝐸 𝑀𝐴𝐼𝑁𝑆 2020…

• July 1, 2020 at 11:06 am

𝑃𝑙𝑧𝑧 𝑢𝑝𝑑𝑎𝑡𝑒 𝐽𝐸𝐸 𝑀𝐴𝐼𝑁𝑆 2020..

• June 30, 2020 at 9:20 pm

• June 23, 2020 at 7:43 pm

🙂

• June 23, 2020 at 12:24 pm

• June 11, 2020 at 6:21 pm

Very useful. Thanks 😊

• May 26, 2020 at 5:35 pm

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• May 25, 2020 at 8:38 am

Plz can update jee mains 2020

• May 11, 2020 at 7:41 pm

enter new mains questions and jee adv questions also

• May 4, 2020 at 12:58 pm

super questions