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)$\left(\frac{5}{4}, 0\right)$

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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

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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]**
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]**
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]**
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) $x^{2}+y^{2}+2 x+2 y-7=0$

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

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**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)

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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

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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
$(4) \pm 2$

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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]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

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

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

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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]**
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]**
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]**
Very nice 👍

It’s like a quize

Very helpful for me

lodu salla kuch bhi question dalte hai madarchod ke bache salle

PLZ UPOLAD RECENTLY QUESTION

These are easy question but very helpful thankyou esaral please upload advance recent questions sir

Upload new questions also

Upload recent questions also..

nice explanation

Hii sir very nice esaral pattern

Thanks sir

Super

These questions are good but you need to update to upload recent years also

thank u for this questions hope this will help in my exams

Questions are lovely

Cheaters never join esaral take coaching

Update 2019 and 2020 questions

Great very helpful

Good Questions.

Really Helpful

𝑈𝑝𝑑𝑎𝑡𝑒 𝐽𝐸𝐸 𝑀𝐴𝐼𝑁𝑆 2020…

𝑃𝑙𝑧𝑧 𝑢𝑝𝑑𝑎𝑡𝑒 𝐽𝐸𝐸 𝑀𝐴𝐼𝑁𝑆 2020..

Please upload jee advanced questions also. Above questions are good. Please upload some difficult and previous 15years questions. Thanks so much

🙂

PLEASE UPLOAD JEE ADVANCED QUESTIONS ALSO

Well done esaral👍👍👍

Very useful. Thanks 😊

2019 k ques kha hai beta?

Plz can update jee mains 2020

enter new mains questions and jee adv questions also

super questions