datavorous/entrance-exam-dataset huggingface dataset:
The equation of plane through the line of intersection of planes $a x+b y+c z+d=0, a^{\prime} x+b^{\prime} y+c^{\prime} z+d^{\prime}=0$ and parallel to the line $y=0, z=0$ is

Options:
A $\left(a b^{\prime}-a^{\prime} b\right) x+\left(b c^{\prime}-b^{\prime} c\right) y+\left(a d^{\prime}-a^{\prime} d\right)=0$
B $\left(a b^{\prime}-a^{\prime} b\right) x+\left(b c^{\prime}-b^{\prime} c\right) y+\left(a d^{\prime}-a^{\prime} d\right) z=0$
C $\left(a b^{\prime}-a^{\prime} b\right) y+\left(a c^{\prime}-a^{\prime} c\right) z+\left(a d^{\prime}-a^{\prime} d\right)=0$
D None of these

The correct answer is:
$\left(a b^{\prime}-a^{\prime} b\right) y+\left(a c^{\prime}-a^{\prime} c\right) z+\left(a d^{\prime}-a^{\prime} d\right)=0$

Solution:
The equation of a plane through the line of intersection of the planes $a x+b y+c z+d=0$ and $a^{\prime} x+b^{\prime} y+c^{\prime} z+d^{\prime}=0$ is
$\begin{aligned}
& (a x+b y+c z+d)+\lambda\left(a^{\prime} x+b^{\prime} y+c^{\prime} z+d^{\prime}\right)=0 \\
& \text { or } x\left(a+\lambda a^{\prime}\right)+y\left(b+\lambda b^{\prime}\right)+z\left(c+\lambda c^{\prime}\right)+d+\lambda d^{\prime}=0
\end{aligned}$
This is parallel to $x$-axis i.e., $y=0, z=0$
$\therefore 1\left(a+\lambda a^{\prime}\right)+0\left(b+\lambda b^{\prime}\right)+0\left(c+\lambda c^{\prime}\right)=0 \Rightarrow \lambda=-\frac{a}{a^{\prime}}$
Putting the value of $\lambda$ in (i), the required plane is
$\begin{aligned}
& y\left(a^{\prime} b-a b^{\prime}\right)+z\left(a^{\prime} c-a c^{\prime}\right)+a^{\prime} d-a d^{\prime}=0 \\
& \text { i.e., }\left(a b^{\prime}-a^{\prime} b\right) y+\left(a c^{\prime}-a^{\prime} c\right) z+a d^{\prime}-a^{\prime} d=0 .
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of projectile is \( \mathrm{y}=\sqrt{3} \mathrm{x}-\frac{\mathrm{g}}{2} \mathrm{x}^{2} \). The angle of projection and initial velocity is -

Options:
A \( 30^{\circ}, 4 \mathrm{~m} / \mathrm{s} \)
B \( 60^{\circ}, 2 \mathrm{~m} / \mathrm{s} \)
C \( 60^{\circ}, 4 \mathrm{~m} / \mathrm{s} \)
D \( 90^{\circ}, 4 \mathrm{~m} / \mathrm{s} \)

The correct answer is:
\( 60^{\circ}, 2 \mathrm{~m} / \mathrm{s} \)

Solution:
Equation of trajectory of a projectile
y=tanâ¡Î¸x-g2u2cos2â¡Î¸x2 .... (i)
Given equation is
y=3 x-g2x2 .... (ii)
Comparing (1) and (2), we get
tanâ¡Î¸=3âÎ¸=60o
And 2u2cos2â¡Î¸=2
â¹u=1cosÎ¸=1cosâ¡60o=2 m/s

Topic: Physics, Motion In Two Dimensions, JEE Main
The equation of radical axis of the circles $2 x^2+2 y^2-7 x=0$ and $x^2+y^2-4 y-7=0$ is

Options:
A $7 x+8 y+14=0$
B $7 x-8 y+14=0$
C $7 x-8 y-14=0$
D None of these

The correct answer is:
$7 x-8 y-14=0$

Solution:
Equation of radical axis, $S_1-S_2=0$
$\begin{aligned} & \text { i.e., }\left(2 x^2+2 y^2-7 x\right)-\left(2 x^2+2 y^2-8 y-14\right)=0 \\ & \Rightarrow-7 x+8 y+14=0, \quad \therefore 7 x-8 y-14=0\end{aligned}$

Topic: Mathematics, Circle, JEE Main
The equation of S.H.M. is $y=a \sin (2 \pi n t+\alpha)$, then its phase at time $t$ is

Options:
A $2 \pi n t$
B $\alpha$
C $2 \pi n t+\alpha$
D $2 \pi t$

The correct answer is:
$2 \pi n t+\alpha$

Solution:
$\begin{aligned} & y=a \sin (2 \pi n+a) \\ & \text { at time 't' } \\ & \text { phase is }(2 \pi n t+a)\end{aligned}$

Topic: Physics, Oscillations, JEE Main
The equation of state for $5 \mathrm{~g}$ of oxygen at a pressure $P$ and temperature $T$, when occupying a volume $V$, will be:
(where $R$ is the gas constant)

Options:
A $P V=(5 / 32) R T$
B $P V=5 R T$
C $P V=(5 / 2) R T$
D $P V=(5 / 16) R T$

The correct answer is:
$P V=(5 / 32) R T$

Solution:
We know that $P V=\underline{n} R T$
$$
\begin{gathered}
\therefore n=\frac{m}{\text { moleculer mass }}=\frac{5}{32} \\
\Rightarrow \quad P V=\left(\frac{5}{32}\right) R T
\end{gathered}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The equation of state for a gas is given by PV = nRT + α V , where n is the number of moles and α is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are T0 and P0 respectively. The work done by the gas when its temperature doubles isobarically will be :

Options:
A P 0 T 0 lnâ¡ 2
B P 0 T 0 R P 0 + α
C P 0 T 0 R P 0 - α
D P 0 T 0 R

The correct answer is:
P 0 T 0 R P 0 - α

Solution:
PV = nRT + α V
P 0 dV dT = R + α dV dT
P 0 - α dV dT = R
∴ dV = RdT P 0 - α
∴ W = ∫ P 0 dV = ∫ T 0 2 T 0 P 0 R P 0 - α dT
= P 0 T 0 R P 0 - α

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2014 (09 Apr Online)
The equation of state for $\mathrm{n}$ moles of an ideal gas is $\mathrm{PV}=\mathrm{nRT}$, where $\mathrm{R}$ is a constant. The SI unit for $\mathrm{R}$ is

Options:
A $\mathrm{JK}^{-1}$ per molecule
B $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
C $\mathrm{J} \mathrm{Kg}^{-1} \mathrm{~K}^{-1}$
D $\mathrm{JK}^{-1} \mathrm{~g}^{-1}$

The correct answer is:
$\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Solution:
$\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The equation of state of a gas is given by $\left(p+\frac{a}{v^{3}}\right)\left(V-b^{2}\right)=c T,$ where $p, V, T$ are pressure, volume and temperature respectively and $a, b, c$ are constants. The dimensions of $a$ and $b$ are respectively

Options:
A $\left[\mathrm{ML}^{8} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3 / 2}\right]$
B $\left[\mathrm{ML}^{5} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3}\right]$
C $\left[\mathrm{ML}^{5} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{6}\right]$
D $\left[\mathrm{ML}^{6} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3 / 2}\right]$

The correct answer is:
$\left[\mathrm{ML}^{8} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3 / 2}\right]$

Solution:
$[P]=\left[\frac{a}{V^{3}}\right] \Rightarrow\left[M L^{-1} T^{-2}\right]=\frac{a}{\left[L^{3}\right]^{3}}$
$\begin{aligned} \Rightarrow \quad a &=\left[M L^{8} T^{-2}\right] \\ &[V]=\left[b^{2}\right] \Rightarrow\left[L^{3}\right]=b^{2} \Rightarrow b=\left[L^{3 / 2}\right] \end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
The equation of state of a real gas is given by P+aV2(V-b)=RT, where P, V and T are pressure, volume and temperature respectively and R is the universal gas constant. The dimensions of ab2 is similar to that of :

Options:
A PV
B P
C RT
D R

The correct answer is:
P

Solution:
From the dimensional analysis, it follows that

[P]=aV2⇒[a]=PV2 ...1

And,

[V]=[b]

Hence,

ab2=PV2V2=[P]

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The equation of state of some gases can be expressed as $\left(P+\frac{a}{V^2}\right)(V-b)=R T$. Here $P$ is the pressure, $V$ is the volume, $T$ is the absolute temperature and $a, b, R$ are constants. The dimensions of ' $a$ ' are

Options:
A $M L^5 T^{-2}$
B $M {L}^{-1} T^{-2}$
C $M^0 L^3 T^0$
D $M^0 L^6 T^0$

The correct answer is:
$M L^5 T^{-2}$

Solution:
By principle of dimensional homogenity
$\begin{aligned}{\left[\frac{a}{V^2}\right] } & {=[P] } \\\therefore & {[a]=[P]\left[V^2\right]=\left[M {L}^{-1} T^{-2}\right] \times\left[L^6\right]=\left[M L^5 T^{-2}\right] }\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
The equation of state or 2g of oxygen at a pressure ‘P’ and temperature ‘T’, when occupying a volume ‘V’ will be

Options:
A pV=16RT
B pV=RT
C pV=116RT
D pV=2RT

The correct answer is:
pV=116RT

Solution:
The ideal gas equation is given by
pV=μRT
where, μ= number of moles in gas
As, μ=GivenmassMolecularmass=mM
Here, m = 2g
And for oxygen O2,M=32g
Substituting these values in Eq. (i), we get
pV=mMRT=332RT
∴pV=116

Topic: Physics, Kinetic Theory of Gases, JEE Main
The equation of straight line passing through the intersection of the lines $x-2 y=1$ and $x+3 y=2$ and parallel to $3 x+4 y=0$ is

Options:
A $3 x+4 y+5=0$
B $3 x+4 y-10=0$
C $3 x+4 y-5=0$
D $3 x+4 y+6=0$

The correct answer is:
$3 x+4 y-5=0$

Solution:
The intersection point of lines $x-2 y=1$ and $x+3 y=2$ is $\left(\frac{7}{5}, \frac{1}{5}\right)$ and the slope of required line
$=-\frac{3}{4}$
$\therefore$ Equation of required line is $y-\frac{1}{5}=\frac{-3}{4}\left(x-\frac{7}{5}\right)$
$\Rightarrow \frac{3 x}{4}+y=\frac{21}{20}+\frac{1}{5} \Rightarrow 3 x+4 y=5 \Rightarrow 3 x+4 y-5=0$

Topic: Mathematics, Straight Lines, JEE Main
The equation of straight line passing through the point $(a, b, c)$ and parallel to $z$ - axis, is

Options:
A $\frac{x-a}{1}=\frac{y-b}{1}=\frac{z-c}{0}$
B $\frac{x-a}{0}=\frac{y-b}{1}=\frac{z-c}{1}$
C $\frac{x-a}{1}=\frac{y-b}{0}=\frac{z-c}{0}$
D $\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$

The correct answer is:
$\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$

Solution:
The line through $(a, b, c)$ is $\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n}$
Since the line is parallel to $z$-axis, therefore any point on this line will be of the form $\left(a, b, c\right)$.
Also any point on line (i) is $(l r+a, m r+b, n r+c)$.
Hence
$\begin{gathered}
l r+a=a \\
m r+b=b
\end{gathered} \Rightarrow l=m=0$
$\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of tangent and normal at point $(3,-2)$ of ellipse $4 x^2+9 y^2=36$ are

Options:
A $\frac{x}{3}-\frac{y}{2}=1, \frac{x}{2}+\frac{y}{3}=\frac{5}{6}$
B $\frac{x}{3}+\frac{y}{2}=1, \frac{x}{2}-\frac{y}{3}=\frac{5}{6}$
C $\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}-\frac{y}{2}=\frac{5}{6}$
D None of these

The correct answer is:
$\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}-\frac{y}{2}=\frac{5}{6}$

Solution:
Given equation of ellipse is $4 x^2+9 y^2=36$ or $\frac{x^2}{9}+\frac{y^2}{4}=1$
Tangent at point $(3,-2)$ is $\frac{(3) x}{9}+\frac{(-2) y}{4}=1$ or $\frac{x}{3}-\frac{y}{2}=1$
$\therefore$ Normal is $\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}=\mathrm{k}$ and it passes through point $(3,-2)$,
Then, $\frac{3}{2}-\frac{2}{3}=k \Rightarrow k=\frac{5}{6}$
Hence, normal is $\frac{x}{2}+\frac{y}{3}=\frac{5}{6}$ and tangent is $\frac{x}{3}-\frac{y}{2}=1$.

Topic: Mathematics, Ellipse, JEE Main
The equation of tangent and normal at point $(3,-2)$ of ellipse $4 x^2+9 y^2=36$ are

Options:
A $\frac{x}{3}-\frac{y}{2}=1, \frac{x}{2}+\frac{y}{3}=\frac{5}{6}$
B $\frac{x}{3}+\frac{y}{2}=1, \frac{x}{2}-\frac{y}{3}=\frac{5}{6}$
C $\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}-\frac{y}{2}=\frac{5}{6}$
D None of these

The correct answer is:
$\frac{x}{3}-\frac{y}{2}=1, \frac{x}{2}+\frac{y}{3}=\frac{5}{6}$

Solution:
Given equation of ellipse is $4 x^2+9 y^2=36$ or $\frac{x^2}{9}+\frac{y^2}{4}=1$
Tangent at point $(3,-2)$ is $\frac{(3) x}{9}+\frac{(-2) y}{4}=1$ or $\frac{x}{3}-\frac{y}{2}=1$
$\therefore$ Normal is $\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}=\mathrm{k}$ and it passes through point $(3,-2)$.
Then, $\frac{3}{2}-\frac{2}{3}=k \Rightarrow k=\frac{5}{6}$
Hence, normal is $\frac{x}{2}+\frac{y}{3}=\frac{5}{6}$ and tangent is $\frac{x}{3}-\frac{y}{2}=1$.

Topic: Mathematics, Ellipse, JEE Main
The equation of tangent to the curve $y=2 \cos x$ at $x=\frac{\pi}{4}$ is

Options:
A $y-\sqrt{2}=2 \sqrt{2}\left(x-\frac{\pi}{4}\right)$
B $y+\sqrt{2}=\sqrt{2}\left(x+\frac{\pi}{4}\right)$
C $y-\sqrt{2}=-\sqrt{2}\left(x-\frac{\pi}{4}\right)$
D $y-\sqrt{2}=\sqrt{2}\left(x-\frac{\pi}{4}\right)$

The correct answer is:
$y-\sqrt{2}=-\sqrt{2}\left(x-\frac{\pi}{4}\right)$

Solution:
$y=2 \cos x$
At $x=\frac{\pi}{4}, y=\frac{2}{\sqrt{2}}=\sqrt{2}$ and $\frac{d y}{d x}=-2 \cdot \sin x$
$\therefore\left(\frac{d y}{d x}\right)_{x=\pi / 4}=-\sqrt{2}$
$\therefore$
Equation of tangent at $\left(\frac{\pi}{4}, \sqrt{2}\right)$ is $y-\sqrt{2}=-\sqrt{2}\left(x-\frac{\pi}{4}\right)$

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of tangent to the curve \(y=\sin x\) at the point \((\pi, 0)\) is

Options:
A \(x+y=0\)
B \(x+y=\pi\)
C \(x-y=\pi\)
D \(x-y=0\)

The correct answer is:
\(x+y=\pi\)

Solution:
\(y=\sin x \Rightarrow \frac{d y}{d x}=\cos x \Rightarrow\left(\frac{d y}{d x}\right)_{(\pi, 0)}=-1\)
Therefore the equation of tangent at \((\pi, 0)\) is given by
\(y-0=-1(x-\pi) \Rightarrow x+y=\pi\)

Topic: Mathematics, Indefinite Integration, JEE Main
The equation of the bisectors of the angles between the lines represented by the equation 2x+22+3x+2y-2-2y-22=0 is

Options:
A 3x2-8xy-3y2-28x+4y+32=0
B 3x2+8xy-3y2+28x-4y+32=0
C 3x2-8xy-3y2+28x-4y+32=0
D 3x2-8xy-3y2+28x-4y-32=0

The correct answer is:
3x2-8xy-3y2+28x-4y+32=0

Solution:
Equation of the bisectors would be
x+22-y-222--2=x+2(y-2)32
3x2+4x+4-y2+4y-4=8 xy-2x+2y-4
3x2+12x-3y2+12y=8xy-16x+16y-32
3x2-8xy-3y2+28x-4y+32=0

Topic: Mathematics, Straight Lines, JEE Main
The equation of the chord AB of the circle x2+y2=2 passing through the point P2, 2 such that PBPA=3 is given by

Options:
A x=3y
B x=y
C y-2=3 x-2
D None of these

The correct answer is:
x=y

Solution:
We know,

PA.PB=PT2=22+22-2=6 …(1)
PBPA=3 …(2)
From (1) and (2), we get,

PA=2 , PB=32
⇒ AB=22 .
Now, the diameter of the circle is 22 (as the radius is 2 )
Hence, the line passes through the center

⇒y=x

Topic: Mathematics, Circle, JEE Main
The equation of the chord of contact, if the tangents are drawn from the point $(5,-3)$ to the circle $x^2+y^2=10$, is

Options:
A $5 x-3 y=10$
B $5 x+3 y=10$
C $3 x+5 y=10$
D $3 x-5 y=10$

The correct answer is:
$5 x-3 y=10$

Solution:
It is $x x_1+y y_1=a^2$ i.e., $5 x-3 y=10$.

Topic: Mathematics, Circle, JEE Main
The equation of the chord of the circle $x^2+y^2=a^2$ having $\left(x_1, y_1\right)$ as its mid-point is

Options:
A $x y_1+y x_1=a^2$
B $x_1+y_1=a$
C $x x_1+y y_1=x_1^2+y_1^2$
D $x x_1+y y_1=a^2$

The correct answer is:
$x x_1+y y_1=x_1^2+y_1^2$

Solution:
$T=S_1$ is the equation of desired chord, hence
$x x_1+y y_1-a^2=x_1^2+y_1^2-a^2 \Rightarrow x x_1+y y_1=x_1^2+y_1^2 \text {. }$

Topic: Mathematics, Circle, JEE Main
The equation of the chord of the ellipse $x^2+4 y^2=4$ having the middle point at $\left(-1, \frac{1}{2}\right)$ is

Options:
A $x-2 y+2=0$
B $x+2 y=0$
C $3 x-2 y+4=0$
D $2 x-2 y+3=0$

The correct answer is:
$x-2 y+2=0$

Solution:
We know that, the equation of the chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose mid-point is $x_1, \quad y_1$ is $T=S_1$ i.e. $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}$. Thus, the chord of the ellipse $x^2+4 y^2=4$ having mid-point at $\left(-1, \frac{1}{2}\right)$ is
$\begin{aligned} & x-1+4 y \times \frac{1}{2}=-1^2+4 \times \frac{1^2}{2} \\ & \Rightarrow-x+2 y=1+4 \times \frac{1}{4} \\ & \Rightarrow-x+2 y=2 \\ & \Rightarrow x-2 y+2=0 .\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
The equation of the circle concentric with the circle $x^2+y^2+8 x+10 y-7=0$ and passing through the centre of the circle $x^2+y^2-4 x-6 y=0$ is

Options:
A $x^2+y^2+8 x+10 y+59=0$
B $x^2+y^2+8 x+10 y-59=0$
C $x^2+y^2-4 x-6 y+87=0$
D $x^2+y^2-4 x-6 y-87=0$

The correct answer is:
$x^2+y^2+8 x+10 y-59=0$

Solution:
Centre is $(-4,-5)$ and passes through $(2,3)$.

Topic: Mathematics, Circle, JEE Main
The equation of the circle described on the chord 3x⁡+y⁡+5=0 of the circle x⁡2+y⁡2=16 as the diameter is

Options:
A x⁡2+y⁡2+3x⁡+y⁡+1=0
B x⁡2+y⁡2+3x⁡+y⁡-22=0
C x⁡2+y⁡2+3x⁡+y⁡-11=0
D x⁡2+y⁡2+3x⁡+y⁡-2=0

The correct answer is:
x⁡2+y⁡2+3x⁡+y⁡-11=0

Solution:
We know that the equation of the family of circle passing through the point of intersection of a circle S=0 and a line L=0 is S+λL=0.

Hence, the family of circle passing through points of intersection of given circle x2+y2-16=0 and chord 3x+y+5=0 is (x2+y2-16)+λ3x+y+5=0

⇒x2+y2+3λx+λy+5λ-16=0

We know that, the centre of a circle x2+y2+2gx+2fy+c=0 is -g, -f

Hence, centre of the circle x2+y2+3λx+λy+5λ-16=0 is -3λ2, -λ2 and since, for this circle AB is diameter, thus -3λ2, -λ2 must lie on AB i.e. 3x+y+5=0

⇒3-3λ2+-λ2 +5=0

⇒-9λ2-λ2+5=0

⇒λ=1.

Therefore, the equation of the required circle is x2+y2+3×1×x+1×y+5×1-16=0

⇒x2+y2+3x+y-11=0.

Topic: Mathematics, Circle, JEE Main, JEE Main 2014 (19 Apr Online)
The equation of the circle having its centre on the line $x+2 y-3=0$ and passing through the points of intersection of the circles $x^2+y^2-2 x-4 y+1=0$ and $x^2+y^2-4 x-2 y+4=0$, is

Options:
A $x^2+y^2-6 x+7=0$
B $x^2+y^2-3 y+4=0$
C $x^2+y^2-2 x-2 y+1=0$
D $x^2+y^2+2 x-4 y+4=0$

The correct answer is:
$x^2+y^2-6 x+7=0$

Solution:
Required circle will be $S_1+\lambda S_2=0, \lambda \neq-1$
i.e., $x^2+y^2-2 x-4 y+1+\lambda\left(x^2+y^2-4 x-2 y+4\right)=0$
$\begin{aligned}& \Rightarrow x^2+y^2-2 \frac{(1+2 \lambda)}{1+\lambda} x-2 \frac{(2+\lambda)}{1+\lambda} y+\frac{1+4 \lambda}{1+\lambda}=0 \\& \text { Its centre }\left(\frac{1+2 \lambda}{1+\lambda}, \frac{2+\lambda}{1+\lambda}\right) \text { lies on } x+2 y-3=0\end{aligned}$
$\therefore \frac{1+2 \lambda}{1+\lambda}+2\left(\frac{2+\lambda}{1+\lambda}\right)-3=0 \Rightarrow \lambda=-2 \text {. }$
$\therefore$ The circle is $x^2+y^2-6 x+7=0$

Topic: Mathematics, Circle, JEE Main
The equation of the circle of radius 5 and touching the co-ordinate axes in third quadrant is

Options:
A $(x-5)^2+(y+5)^2=25$
B $(x+5)^2+(y+5)^2=25$
C $(x+4)^2+(y+4)^2=25$
D $(x+6)^2+(y+6)^2=25$

The correct answer is:
$(x+5)^2+(y+5)^2=25$

Solution:
Since the circle touches the coordinate axes in third quadrant therefore.
centre of the circle is $(-5,-5)$ Then equation of circle is, $(x+5)^2+(y+5)^2=25$.

Topic: Mathematics, Circle, JEE Main
The equation of the circle passing through the foci of the ellipse x216+y29=1, and having centre at 0,3 is

Options:
A x2+y2-6y-5=0
B x2+y2-6y+5=0
C x2+y2-6y-7=0
D x2+y2-6y+7=0

The correct answer is:
x2+y2-6y-7=0

Solution:
For x216+y29=1

ae=16-9=7
Foci=±7,0
Radius=7+9=4
∴Equation of circle=x-02+y-32=42
⇒x2+y2-6y-7=0

Topic: Mathematics, Ellipse, JEE Main
The equation of the circle passing through the foci of the ellipse x216+y29=1, and having centre at 0,3 is

Options:
A x2+y2-6y-5=0
B x2+y2-6y+5=0
C x2+y2-6y-7=0
D x2+y2-6y+7=0

The correct answer is:
x2+y2-6y-7=0

Solution:
For x216+y29=1

ae=16-9=7
Foci=±7,0
Radius=7+9=4
∴Equation of circle=x-02+y-32=42
⇒x2+y2-6y-7=0

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2013 (07 Apr)
The equation of the circle passing through the origin and cutting intercepts of length 3 and 4 units from the positive axes, is

Options:
A $x^2+y^2+6 x+8 y+1=0$
B $x^2+y^2-6 x-8 y=0$
C $x^2+y^2+3 x+4 y=0$
D $x^2+y^2-3 x-4 y=0$

The correct answer is:
$x^2+y^2-3 x-4 y=0$

Solution:
Obviously the centre of the circle is $\left(\frac{3}{2}, 2\right)$
Therefore, the equation of circle is $\left(x-\frac{3}{2}\right)^2+(y-2)^2=\left(\frac{5}{2}\right)^2 \Rightarrow x^2+y^2-3 x-4 y=0$

Topic: Mathematics, Circle, JEE Main
The equation of the circle passing through the point $(1,2)$ and through the points of intersection of $x^2+y^2-4 x-6 y-21=0$ and $3 x+4 y+5=0$ is given by

Options:
A
$x^2+y^2+2 x+2 y+11=0$

B
$x^2+y^2-2 x+2 y-7=0$

C
$x^2+y^2+2 x-2 y-3=0$

D
$x^2+y^2+2 x+2 y-11=0$

The correct answer is:
$x^2+y^2+2 x+2 y-11=0$

Solution:
Point $(1,2)$ lies on the circle $x^2+y^2+2 x+2 y$ $-11=0$, because coordinates of point $(1,2)$ satisfy the equation
$$
x^2+y^2+2 x+2 y-11=0
$$
Now, $x^2+y^2-4 x-6 y-21=0 \quad$...(i)
$$
\begin{aligned}
& x^2+y^2+2 x+2 y-11=0 \\
& 3 x+4 y+5=0
\end{aligned}
$$
From (i) and (iii),
$$
\begin{array}{r}
x^2+\left(-\frac{3 x+5}{4}\right)^2-4 x-6\left(-\frac{3 x+5}{4}\right)-21=0 \\
\Rightarrow \quad 16 x^2+9 x^2+30 x+25-64 x \\
+72 x+120-336=0
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow \quad 25 x^2+38 x-191=0 \\
& \text { From (ii) and (iii), } \\
& x^2+\left(-\frac{3 x+5}{4}\right)^2+2 x+2\left(-\frac{3 x+5}{4}\right)-11=0 \\
& \Rightarrow \quad 16 x^2+9 x^2+30 x+25+32 x
\end{aligned}
$$
Thus we get the same equation from (ii) and (iii) as we get from equation (i) and (iii). Hence the point of intersections of (ii) and (iii) will be same as the point of intersections of(i) and (iii). Therefore the circle (ii) passing through the point of intersection of circle(i) and point $(1,2)$ also as shown in the figure.
Hence equation(ii) i.e. $x^2+y^2+2 x+2 y-11=0$ is the equation of required circle.

Topic: Mathematics, Circle, JEE Main, JEE Main 2012 (07 May Online)
The equation of the circle which passes through the origin and cuts off intercepts of 2 units length from negative coordinate axes, is

Options:
A $x^2+y^2-2 x+2 y=0$
B $x^2+y^2+2 x-2 y=0$
C $x^2+y^2+2 x+2 y=0$
D $x^2+y^2-2 x-2 y=0$

The correct answer is:
$x^2+y^2+2 x+2 y=0$

Solution:
Since the circle passes through $(0,0)$, hence $c=0$. Also $2 \sqrt{g^2-c}=2 \Rightarrow g=1$ and $2 \sqrt{f^2-c}=2 \Rightarrow f=1$
Hence radius is $\sqrt{2}$ and centre is $(-1,-1)$. Therefore, the required equation is $x^2+y^2+2 x+2 y=0$
Trick : Obviously the centre of circle lies in III quadrant, which is given by (3).

Topic: Mathematics, Circle, JEE Main
The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^{2}+y^{2}-6 x+8=0$ and $x^{2}+y^{2}-2 x-2 y=7$ is

Options:
A $3 x^{2}+3 y^{2}-8 x-13 y=0$
B $3 x^{2}+3 y^{2}-8 x+29 y=0$
C $3 x^{2}+3 y^{2}+8 x+29 y=0$
D $3 x^{2}+3 y^{2}-8 x-29 y=0$

The correct answer is:
$3 x^{2}+3 y^{2}-8 x+29 y=0$

Solution:
Let the required equation of circle be
$$
x^{2}+y^{2}+2 g x+2 f y=0
$$
The above circle cuts the given circles orthogonally.
$$
\therefore 2(-3 g)+2 f(0)=8 \Rightarrow 2 g=-\frac{8}{3}
$$
and $-2 g-2 f=-7$ $\Rightarrow 2 f=+7+\frac{8}{3}=\frac{29}{3}$
$\therefore$ Required equation of circle is $x^{2}+y^{2}-\frac{8}{3} x+\frac{29}{3} y=0$
or $3 x^{2}+3 y^{2}-8 x+29 y=0$

Topic: Mathematics, Circle, JEE Main
The equation of the circle which passes through the points $(1,0),(0,-6)$ and $(3,4)$ is

Options:
A $4 x^{2}+4 y^{2}+142 x+47 y+140=0$
B $4 x^{2}+4 y^{2}-142 x-47 y+138=0$
C $4 x^{2}+4 y^{2}-142 x+47 y+138=0$
D $4 x^{2}+4 y^{2}+150 x-49 y+138=0$

The correct answer is:
$4 x^{2}+4 y^{2}-142 x+47 y+138=0$

Solution:
Let $\mathrm{A}=(1,0), \mathrm{B}=(0,-6), \mathrm{C}=(3,4)$
Equation of $\mathrm{AB}$ is $\mathrm{L}: \frac{\mathrm{y}-0}{-6-0}=\frac{\mathrm{x}-1}{0-1}$
$\Rightarrow \frac{y}{-6}=\frac{x-1}{-1} \Rightarrow y=6 x-6$
$\Rightarrow 6 x-y-6=0 .$
Equation of circle (c) with $\mathrm{AB}$ as diameter is $(\mathrm{x}-1)(\mathrm{x}-$ $0)+(y-0)(y+6)=0$
$\Rightarrow x^{2}-x+y^{2}+6 y=0 .$
The system of circle passing through the intersection of the circle $\mathrm{C}$ and the line $\mathrm{L}$ is given by $\mathrm{C}+\mathrm{kL}=0$ $\Rightarrow x^{2}-x+y^{2}+6 y+k(6 x-y-6)=0$
This circle is passing through $(3,4)$. $\therefore(3)^{2}-3+(4)^{2}+6(4)+\mathrm{k}[6(3)-4-6]=0$
$\Rightarrow 9-3+16+24+\mathrm{k}(18-10)=0$
$\Rightarrow 46+8 \mathrm{k}=0 \Rightarrow 8 \mathrm{k}=-46 \Rightarrow \mathrm{k}=\frac{-46}{8}=\frac{-23}{4}$
$\therefore$ Equation of circle is
$x^{2}-x+y^{2}+6 y+\left(\frac{-23}{4}\right)(6 x-y-6)=0$
$\Rightarrow 4 x^{2}-4 x+4 y^{2}+24 y-138 x+23 y+138=0$
$\Rightarrow 4 x^{2}+4 y^{2}-142 x+47 y+138=0 .$

Topic: Mathematics, Circle, JEE Main
The equation of the circle which passes through the points $(2,3)$ and $(4,5)$ and the centre lies on the straight line $y-4 x+3=0$, is

Options:
A $x^2+y^2+4 x-10 y+25=0$
B $x^2+y^2-4 x-10 y+25=0$
C $x^2+y^2-4 x-10 y+16=0$
D $x^2+y^2-14 y+8=0$

The correct answer is:
$x^2+y^2-4 x-10 y+25=0$

Solution:
First find the centre. Let centre be $(h, k)$, then
$\sqrt{(h-2)^2+(k-3)^2}=\sqrt{(h-4)^2+(k-5)^2}...(i)$
and $k-4 h+3=0...(ii)$
From (i), we get $-4 h-6 k+8 h+10 k=16+25-4-9$
or $4 h+4 k-28=0$ or $h+k-7=...(iii)$
From (iii) and (ii), we get $(h, k)$ as $(2,5)$. Hence centre is $(2,5)$ and radius is 2 . Now find the equation of circle.
Trick : Obviously, circle ${x^2+y^2-4 x-10 y+25=0}$ passes through $(2,3)$ and $(4,5)$

Topic: Mathematics, Circle, JEE Main
The equation of the circle which passes through the points A0,5 and B6,1 and whose centre lies on the line 12x+5y=25 is

Options:
A 3x2+3y2+10x+6y+15=0
B 3x2+3y2-10x-6y-45=0
C x2+y2-6x-6y+5=0
D x2+y2-4x-3y-10=0

The correct answer is:
3x2+3y2-10x-6y-45=0

Solution:
Centre C of the required circle lies on the perpendicular bisector of the line segment joining 0,5 & 6,1
the midpoint of line segment joining A & B is 3,3=D
Slope of AB is -23
Slope of DC is 32
â equation of DC is 3x-2y-3=0
C is the point of intersection of 3x-2y-3=0 and 12x+5y=25âC=53,1
Radius =AC=259+16=133
â equation of the circle is x-532+y-12=1332
â3x2+3y2-10x-6y-45=0

Topic: Mathematics, Circle, JEE Main