datavorous/entrance-exam-dataset huggingface dataset:
If the axes are rotated anticlockwise through an angle $90^{\circ}$, then the equation $x^2=4 \alpha y$ is changed to the equation

Options:
A $y^2=4 a x$
B $x^2=-4 a y$
C $y^2=-4 a x$
D $x^2=4 a y$

The correct answer is:
$y^2=4 a x$

Solution:
Given, equation of parabola
$$
x^2=4 a y
$$
Then,
On rotating the axes anti-clockwise through angle $90^{\circ}$, $X$-axis becomes $Y$-axis and $Y$-axis becomes $X$-axis.
Therefore, the equation of new parabola after rotation is $y^2=4 a x$.

Topic: Mathematics, Parabola, JEE Main
If the axes are rotated through an angle $45^{\circ}$ in the positive direction without changing the origin, then the co-ordinates of the point $(\sqrt{2}, 4)$ in the old system are

Options:
A $(1-2 \sqrt{2}, 1+2 \sqrt{2})$
B $(1+2 \sqrt{2}, 1-2 \sqrt{2})$
C $(2 \sqrt{2}, \sqrt{2})$
D $(\sqrt{2}, 2)$

The correct answer is:
$(1-2 \sqrt{2}, 1+2 \sqrt{2})$

Solution:
If $\theta$ is the angle of rotation, then the co-ordinates in the new system are $x^{\prime}=x \cos \theta+y \sin \theta$, $y^{\prime}=y \cos \theta-x \sin \theta$
Given that $x^{\prime}=\sqrt{2}, y^{\prime}=4$
Thus, $x \cos \theta+y \sin \theta=\sqrt{2}$
$$
y \cos \theta-x \sin \theta=4
$$
Also, $\theta=\frac{\pi}{4} \Rightarrow x \cos \frac{\pi}{4}+y \sin \frac{\pi}{4}=\sqrt{2}$
and $y \cos \frac{\pi}{4}-x \sin \frac{\pi}{4}=4$
$$
\begin{array}{ll}
\Rightarrow & x+y=2 \\
\text { and } & y-x=4 \sqrt{2}
\end{array}
$$
On adding Eqs. (i) and (ii), we get $2 y=2+4 \sqrt{2}$ $\Rightarrow \quad y=1+2 \sqrt{2}$
On subtracting (i) and (ii), we get
$$
\begin{array}{rlrl}
& & 2 x & =2-4 \sqrt{2} \\
& x & =1-2 \sqrt{2}
\end{array}
$$
$$
\Rightarrow \quad x=1-2 \sqrt{2}
$$
Thus, the co-ordinates of $(\sqrt{2}, 4)$ in the old system $(1-2 \sqrt{2}, 1+2 \sqrt{2})$

Topic: Mathematics, Straight Lines, JEE Main
If the axes are rotated through an angle 45°, the coordinates of the point (22,-32) in the new system are

Options:
A (33,−5)
B (−1,−5)
C (53,−7)
D (7,−3)

The correct answer is:
(−1,−5)

Solution:
If the coordinate axes are rotated through an angle θ, then a point P(x, y) will have co-ordinate P(x', y')

Where x'=xcosθ+ysinθ and y'=ycosθ-xsinθ

Here, x, y=22, -32 and θ=45°

Now, x'=22cos45°-32sin45°

⇒x'=22×12-32×12

⇒x'=2-3=-1

And, y'=-32cos45°-22sin45°

y'=-32×12-22×12

⇒y'=-3-2=-5

Therefore the coordinates of the given point in the new system is -1, -5

Topic: Mathematics, Straight Lines, JEE Main
If the axes are shifted to the point $(1,-2)$ with out solution, then the equation $2 x^{2}+y^{2}-4 x+4 y=0$ becomes

Options:
A $2 X^{2}+3 Y^{2}=6$
B $2 X^{2}+Y^{2}=6$
C $X^{2}+2 Y^{2}=6$
D None of these

The correct answer is:
$2 X^{2}+Y^{2}=6$

Solution:
Substituting $x=X+1$ and $y=Y-2$ in given equation, we get
$$
\begin{array}{l}
2(X+1)^{2}+(Y-2)^{2}-4(X+1)+4(Y-2)=0 \\
\Rightarrow 2 X^{2}+Y^{2}=6
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
If the binding energy of ground state electron in a hydrogen atom is 13.6 eV, then, the energy required to remove the electron from the second excited state of Li 2+ will be: x×10-1 eV. The value of x is _____.

Options:

The correct answer is:
136

Solution:
Given:

Binding energy, EH=E0=13.6 eV.

Now for hydrogen like atom, we can write

ELi2+=E0Z2n2=13.6×99=13.6 eV

=136×10-1 eV

Therefore, x=136.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (31 Jan Shift 2)
If the binding energy of N14 is 7.5 MeV per nucleon and that of N15 is7.7 MeV per nucleon. then the energy required to remove a neutron from N15 is

Options:
A 5.25 MeV
B 0.2 MeV
C 10.5 MeV
D 0.4 MeV

The correct answer is:
10.5 MeV

Solution:
The total binding energy,

N14=7.5×14 MeV

N15=7.7×15 MeV

The energy required to remove a neutron from N15,

E=BE N15-BE N14

E=7.7×15-7.5×14 MeV

⇒ E=10.5 MeV

Topic: Physics, Nuclear Physics, JEE Main
If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is :

Options:
A 13.6 eV
B 3.4 eV
C 122.4 eV
D 30.6 eV

The correct answer is:
30.6 eV

Solution:
Energy of an electron in nth orbit of an ion with atomic number z is, E=-13.6×z2n2 eV
for 1st excited state of Li++, z=3, n=2
E=-13.6×32n2=-13.6×94 eV
= -30.6 eV
∴ Required energy =30.6 eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2014 (09 Apr Online)
If the binding energy of the electron in a hydrogen atom is $13.6 \mathrm{eV}$, the energy required to remove the electron from the first excited state of $\mathrm{Li}^{++}$is

Options:
A
$30.6 \mathrm{eV}$

B
$13.6 \mathrm{eV}$

C
$3.4 \mathrm{eV}$

D
$122.4 \mathrm{eV}$

The correct answer is:
$13.6 \mathrm{eV}$

Solution:
No Solution Available

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2003
If the binding energy per nucleon in ${ }_3^7 \mathrm{Li}$ and ${ }_2^4 \mathrm{He}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$ respectively, then in the reaction
$\mathrm{p}+{ }_3^7 \mathrm{Li} \rightarrow 2{ }_2^4 \mathrm{He}$
energy of proton must be

Options:
A
$39.2 \mathrm{MeV}$

B
$28.24 \mathrm{MeV}$

C
$17.28 \mathrm{MeV}$

D
$1.46 \mathrm{MeV}$

The correct answer is:
$17.28 \mathrm{MeV}$

Solution:
$=(8 \times 7.06-7 \times 5.60) \mathrm{MeV}=17.28 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2006
If the binomial expansion of \( (a+b x)^{-2} \) is \( \frac{1}{4}-3 x+\ldots \), then the values of \( a \) and \( b \) are

Options:
A \( 2,12 \)
B \( 2,10 \)
C \( 1,12 \)
D \( 2,6 \)

The correct answer is:
\( 2,12 \)

Solution:
We have

a+bx-2=a-21+bxa-2

=a-21-2bxa+3b2x2a2-....

=1a2-2bxa3+3b2x2a4-… ....i

We have

a+bx-2=14-3x+... ....ii

Comparing i and ii, we get

1a2=14⇒a=2

And,

2ba3=3

⇒2b=323⇒b=12

Topic: Mathematics, Binomial Theorem, JEE Main
If the body is moving in a circle of radius $r$ with a constant speed $v$, its angular velocity is

Options:
A $v^2 / r$
B $v r$
C $v / r$
D $r / v$

The correct answer is:
$v / r$

Solution:
Velocity of the revolving body $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{w}} \times \overrightarrow{\mathrm{r}}$
$\Rightarrow|\overrightarrow{\mathrm{v}}|=|\overrightarrow{\mathrm{w}}||\overrightarrow{\mathrm{r}}| \sin \theta$ where $\theta$ is the angle between the $\overrightarrow{\mathrm{w}}$ and $\overrightarrow{\mathrm{r}}$
As $\theta=90^{\circ} \Rightarrow \mathrm{v}=$ wrsin $90^{\circ}$
$\mathrm{v}=\mathrm{wr} \times 1 \Rightarrow \mathrm{w}=\frac{\mathrm{v}}{\mathrm{r}}$

Topic: Physics, Motion In Two Dimensions, JEE Main
If the boiling point of H2O is 373 K, and the boiling point of H2S will be :

Options:
A less than 300 K
B equal to 373 K
C more than 373 K
D greater than 300 K but less than 373 K

The correct answer is:
less than 300 K

Solution:
At room temperature water is liquid and has boiling points 373 K due to hydrogen bonding. Where as H2S is gas and it has no hydrogen bonding. Hence boiling point of H2S is less than 300 K [Boiling point of H2S is -60 °C]

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (03 Sep Shift 1)
If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio 2: 1 and their enthalpy of vaporizations are in the ratio 1: 2, then the boiling point elevation constant of X is m times the boiling point elevation constant of Y. The value of m is (nearest integer).

Options:

The correct answer is:
8

Solution:
The boiling point elevation constant Kb of a volatile liquid is given by the following expression
Kb=RTb2MΔHv
Where Tb=boiling point of the liquid
M=molar mass of the liquid
ΔHv= Enthalpy of vaporisation

The boiling points of the two solvents are in the ratio 2:1, which means that the boiling point elevation constants are also in the ratio 2:1. Let's call the boiling point elevation constant of solvent X as (Tb)x, and the boiling point elevation constant of solvent Y as (Tb)y.
Given: Mx=My
Tbx=2Tby
2ΔHvx=ΔHvy
∴ KbxKby=212×21=81=m
∴ m=8

Therefore, the value of m is 8(nearest integer).

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (08 Apr Shift 2)
If the bond dissociation energies of $X Y, X_2$ and $Y_2$ (all diatomic molecules) are in the ratio of 1:1:0.5 and $\Delta_t \mathrm{H}$ for the formation of $X Y$ is $-200 \mathrm{~kJ} \mathrm{~mole}^{-1}$. The bond dissociation energy of $\mathrm{X}_2$ will be

Options:
A
$100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$200 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$300 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
None of these

The correct answer is:
None of these

Solution:
$$
\begin{aligned}
& X Y \longrightarrow X_{(g)}+Y_{(g)} ; \quad \Delta \mathrm{H}=+\mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots . .(\mathrm{i}) \\
& \mathrm{X}_2 \longrightarrow 2 \mathrm{X} ; \quad \Delta \mathrm{H}=+\mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots \ldots \text { (ii) } \\
& \mathrm{Y}_2 \longrightarrow 2 \mathrm{Y} ; \quad \Delta \mathrm{H}=+0.5 \mathrm{a} \mathrm{kJ} / \mathrm{mole} \ldots \ldots \ldots \ldots . \text { (iii) } \\
& \frac{1}{2} \times\left. \text { (ii) }+\frac{1}{2} \times \text { (iii) }-\right.\text { (i), Gives } \\
& \frac{1}{2} \mathrm{X}_2+\frac{1}{2} \mathrm{Y}_2 \longrightarrow \mathrm{XY} ; \quad \Delta \mathrm{H}=\left(+\frac{\mathrm{a}}{2}+\frac{0.5}{2} \mathrm{a}-\mathrm{a}\right) \mathrm{kJ} / \mathrm{mole} \\
& +\frac{\mathrm{a}}{2}+\frac{0.5 \mathrm{a}}{2}-\mathrm{a}=-200 \\
& \mathrm{a}=800 .
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2005
If the Boolean expression (p∧q)⊛(p⊗q) is a tautology, then ⊛ and ⊗ are respectively given by

Options:
A →,→
B ∧,∨
C ∨,→
D ∧,→

The correct answer is:
→,→

Solution:
Option A

(p∧q)→(p→q)

We know that a→b=~a∨b

=(p∧q)⟶(~p∨q)

=~(p∧q)∨(~p∨q)

Using De Morgan's Law i.e. ~a∧b=~a∨~b,

=(~p∨~q)∨(~p∨q)

Using associative law and commutative law i.e. a∨b∨c=a∨b∨c & a∨b=b∨a,

=~p∨(~q∨q)

=~p∨t

=t

Option (B)

(p∧q)∧(p∨q)=(p∧q)

Which is not a tautology.

Option (C)

(p∧q)∨(p→q)

=(p∧q)∨(~p∨q)

=~p∨q

Which is not a tautology.

Option (D)

(p∧q)∧(p→q)

=(p∧q)∧(~p∨q)

=p∧q

Which is not a tautology.

Hence, the correct answer is option A.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (17 Mar Shift 2)
If the Boolean expression p⊕q∧~p⊙q is equivalent to p∧q, where ⊕,⊙∈∧,∨, then the ordered pair ⊕,⊙ is

Options:
A ∨,∧
B ∧,∧
C ∨,∨
D ∧,∨

The correct answer is:
∧,∨

Solution:
1) p∨q∧~p∧q

≡p∨q∧~p∧q (Using associative property)

≡q∧~p∧q

≡~p∧q

2) =p∧q∧~p∧q

≡p∧~p∧q

≡F∧q≡F

3) p∨q∧~p∨q

≡p∧~p∨q

⇒F∨q≡q

4) p∧q∧~p∨q

≡q∧p∧~p∨q

≡q∧p∧~p∨q (using associative property)

≡q∧p∧q

≡p∧q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If the Boolean expression (p⇒q)⇔(q*(~p)) is a tautology, then the Boolean expression p*(~q) is equivalent to:

Options:
A q⇒p
B ~q⇒p
C p⇒~q
D p⇒q

The correct answer is:
q⇒p

Solution:
∵p→q≡~p∨q

So, *≡v

Thus, p*(~q)≡pv(~q)

≡q→p

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If the capacitance of a nanocapacitor is measured in terms of a unit u, made by combining the electronic charge e, Bohr radius a0, Planck's constant h and speed of light c then

Options:
A u=e2a0hc
B u=hce2a0
C u=e2cha0
D u=e2hca0

The correct answer is:
u=e2a0hc

Solution:
∵ Capacitance C=Q∆V
Also, hcλ= hca0= Energy.
∴ C= Q∆V= Q Q∆V Q
∵ W=q∆V ⇒ Q ∆V= Energy
∴ C= Q2Energy= Q2 a0hc
∴ Capacitance= Q2 a0hc
⇒ u= e2 a0hc

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2015 (10 Apr Online)
If the center and radius of the circle z-2z-3=2 are respectively α,β and γ, then 3α+β+γ is equal to

Options:
A 11
B 9
C 10
D 12

The correct answer is:
12

Solution:
Given:

z-2z-3=2

Let z=x+iy, then we have

x-2+iyx-3+iy=2

⇒x-22+y2=2x-32+y2

⇒x2+y2-4x+4=4x2+4y2-24x+36

⇒3x2+3y2-20x+32=0

⇒x2+y2-203x+323=0

So, α,β≡103,0

And,

γ=1009-323=49=23

Therefore,

3α+β+γ=3103+23=12

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (01 Feb Shift 1)
If the center, one of the foci and semi-major axis of an ellipse be $(0,0),(0,3)$ and 5 then its equation is

Options:
A $\frac{x^2}{16}+\frac{y^2}{25}=1$
B $\frac{x^2}{25}+\frac{y^2}{16}=1$
C $\frac{x^2}{9}+\frac{y^2}{25}=1$
D None of these

The correct answer is:
$\frac{x^2}{16}+\frac{y^2}{25}=1$

Solution:
Centre $(0,0)$, focus $(0,3), b=5$
Focus $(0,3) \Rightarrow b e=3 \Rightarrow e=3 / 5 \Rightarrow a=b \sqrt{1-e^2}=4$
Hence the required equation is $\frac{x^2}{16}+\frac{y^2}{25}=1$.

Topic: Mathematics, Ellipse, JEE Main
If the central portion of a convex lens is wrapped in black paper as shown in the figure

Options:
A No image will be formed by the remaining portion of the lens
B The full image will be formed but it will be less bright
C The central portion of the image will be missing
D There will be two images each produced by one of the exposed portions of the lens

The correct answer is:
The full image will be formed but it will be less bright

Solution:
Since aperture of lens reduces so brightness will reduce but their will be no effect on size of image.

Topic: Physics, Ray Optics, JEE Main
If the charge on a capacitor is increased by 2C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

Options:
A 10
B 20
C 30
D 40

The correct answer is:
10

Solution:
Energy stored in a capacitor is given by E=Q22C.

Let initial charge be Q0.

Then, 1.44Q022C=Q0+222C

⇒1.2Q0=Q0+2

Q0=20.2=10 C

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (24 Jun Shift 2)
If the chord of contact of the tangents from the point α,β to the circle x2+y2=r12 is a tangent to the circle x-a2+y-b2=r22, then

Options:
A r22α2+β2=r12-aα-bβ2
B r22α2+β2=r12+aα-bβ2
C r22α2+β2=r12-aα+bβ2
D r22α2+β2=r12+aα+bβ2

The correct answer is:
r22α2+β2=r12-aα-bβ2

Solution:
The equation of the chord of contact of tangents drawn from a point α,β to the circle x2+y2=r12 is
αx+βy=r12
This line touches the circle x-a2+y-b2=r22
∴aα+bβ-r12α2+β2=r2
⇒r22α2+β2=aα+bβ-r122
⇒r22α2+β2=r12-aα-bβ2

Topic: Mathematics, Circle, JEE Main
If the chord $y=m x+1$ of the circle $x^2+y^2=1$ subtends an angle of measure $45^0$ at the major segment of the circle then value of $m$ is

Options:
A
$2 \pm \sqrt{2}$

B
$-2 \pm \sqrt{2}$

C
$-1 \pm \sqrt{2}$

D
none of these

The correct answer is:
$-1 \pm \sqrt{2}$

Solution:
Equation of circles $x^2+y^2=1=(1)^2$
$$
\begin{aligned}
& \Rightarrow x^2+y^2=(y-m x)^2 \Rightarrow x^2=m^2 x^2-2 m x y \Rightarrow x^2\left(1-m^2\right)+2 m x y=0 \\
& \tan 45=\pm \frac{2 \sqrt{m^2-0}}{1-m^2}=\frac{\pm 2 m}{1-m^2} \Rightarrow 1-m^2=\pm 2 m \Rightarrow m^2 \pm 2 m-1=0 \\
& \Rightarrow m=\frac{-2 \pm \sqrt{4+4}}{2}=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2}
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2002
IF THE CHORDS OF CONTACT OF TANGENTS FROM TWO POINTS $\left(x_1, y_1\right)$ AND $\left(x_2, y_2\right)$ TO THE ELLIPSE $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1$ ARE AT RIGHT ANGLES, THEN $\frac{x_1 x_2}{y_1 y_2}$ IS EQUAL TO

Options:
A $\frac{a^2}{b^2}$
B $-\frac{b^2}{a^2}$
C $-\frac{a^4}{b^4}$
D $-\frac{b^4}{a^4}$

The correct answer is:
$-\frac{a^4}{b^4}$

Solution:
Correct Answer
$-\frac{a^4}{b^4}$
Chords of contact are $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$ and $\frac{x x_2}{a^2}+\frac{y y_2}{b^2}=1$
Product of the slopes $=-1$
$\Rightarrow-\frac{x_1 / a^2}{y_1 / b^2} \cdot-\frac{x_2 / a^2}{y_2 / b^2}=-1 \Rightarrow \frac{x_1 x_2}{a^4} \cdot \frac{b^4}{y_1 y_2}=-1 \Rightarrow \frac{x_1 x_2}{y_1 y_2}=-\frac{a^4}{b^4}$

Topic: Mathematics, Ellipse, JEE Main
If the circle \( \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}+2 \mathrm{ky}+6=0 \) and \( \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{ky}+\mathrm{k}=0 \) intersect orthogonally, then \( \mathrm{k} \) is

Options:
A \( 2 \) or \( -3 / 2 \)
B \( -2 \) or \( -3 / 2 \)
C \( 2 \) or \( 3 / 2 \)
D \( -2 \) or \( 3 / 2 \)

The correct answer is:
\( 2 \) or \( -3 / 2 \)

Solution:
Since, the given circles intersect orthogonally.

210+2kk=6+k ∵ 2g1g2+2f1f2=c1+c2
⇒ 2k2-k-6=0 ⇒ k=-32 or 2

Topic: Mathematics, Circle, JEE Main
If the circle whose diameter is the major axis of the ellipse x2a2+y2b2=1 a>b>0 meets the minor axis at point P and the orthocentre of âPF1F2 lies on the ellipse, where F1 and F2 are the foci of the ellipse, then the square of the eccentricity of the ellipse is

Options:
A 5-12
B 3-1
C 12
D 32

The correct answer is:
5-12

Solution:
mF1B⋅mPF2=−1⇒a-00-ae×b-00--ae=-1⇒b2a2=e2⇒e4+e2−1=0
∴e2=5-12

Topic: Mathematics, Ellipse, JEE Main
If the circle x2+y2-2gx+6y-19c=0, g,c∈ℝ passes through the point 6,1 and its centre lies on the line x-2cy=8, then the length of intercept made by the circle on x-axis is

Options:
A 11
B 4
C 3
D 223

The correct answer is:
223

Solution:
Given circle x2+y2-2gx+6y-19c=0

Also given circle passes through 6,1

So, 12g+19c=43 ⋯1

And centre g,-3 lies on x-2cy=8,

So, g+6c=8 …2

On solving equation 1 and 2 we get, c=1 and g=2

Now equation of circle will be, x2+y2-4x+6y-19=0

Now length of intercept on x-axis is given by =2g2-C=222+19=223

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (27 Jul Shift 1)
If the circle x2+y2+2kx+4y-4=0 has it's centre in 4th quadrant and touches the circle x2+y2+6x-2y+6=0 then, k=

Options:
A -5
B -157
C -235
D -1

The correct answer is:
-1

Solution:
∙ x2+y2+2kx+4y-4=0

Centre C1=-k,-2

Radius R1=k2+4+4=k2+8

∙ x2+y2+6x-2y+6=0

Centre C2=-3,1

Radius R2=9+1-6=2

∙ Circle touch each others

C1C2=R1+R2

⇒-3+k2+1+22=k2+8+2

⇒k2-6k+18=k2+8+2

⇒k2-6k+18=k2+8+4+4k2+8

⇒6-6k=4k2+8

⇒31-k=2k2+8

⇒9k2-2k+1=4k2+32

⇒5k2-18k-23=0

⇒k+15k-23=0

⇒k=-1,235

As centre lies in 4th quadrant -k>0

⇒k=-1

Topic: Mathematics, Circle, JEE Main
If the circle x2+y2=4x+8y+5 intersects the line 3x-4y=m at two distinct points, then the number of possible integral values of m is equal to

Options:
A 51
B 50
C 49
D 48

The correct answer is:
49

Solution:
The equation of the circle is x2+y2-4x-8y-5=0
Its centre is 2,4 and radius is 4+16+5=5 units
It the circle intersects the line 3x-4y=m at two distinct points, then the length of the perpendicular from the centre is less than the radius
i.e., 6-16-m5<5
⇒10+m<25⇒-25<m+10<25⇒-35<m<15

Hence, the total number of integral values of m is 49

Topic: Mathematics, Circle, JEE Main
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.

Options:
A
0

B
$\pm 4$

C
$\pm 2$

D
$\pm 3$

The correct answer is:
$\pm 4$

Solution:
$$
\begin{aligned}
& x^2+y^2-6 x-8 y+\left(25-a^2\right)=0 \\
& \text { Radius }=4=\sqrt{9+16+\left(25-a^2\right)} \\
& \Rightarrow \mathrm{a}=\pm 4
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2013 (23 Apr Online)
If the circle $x^{2}+y^{2}=a^{2}$ intersects the hyperbola $x y=c^{2}$ in four points $\left(x_{1}, y_{1}\right)$ for $i=1,2,3$ and 4, then $y_{1}+y_{2}+y_{3}+y_{4}$ equals

Options:
A 0
B $c$
C $a$
D $c^{4}$

The correct answer is:
0

Solution:
$$
\begin{array}{l}
x^{2} y^{2}=c^{4} \\
\Rightarrow y^{2}\left(a^{2}-y^{2}\right)=c^{4} \\
\Rightarrow y^{4}-a^{2} y^{2}+c^{4}=0
\end{array}
$$
Let $y_{1}, y_{2}, y_{3}$ and $y_{4}$ are the roots.
$$
\therefore y_{1}+y_{2}+y_{3}+y_{4}=0
$$

Topic: Mathematics, Hyperbola, JEE Main
If the circle x-a2+y2=25 intersects the circle x2+(y-b)2=16 in such a way that the length of the common chord is 8 units, then the value of a2+b2 is

Options:

The correct answer is:
9

Solution:
The equation of the common chord is -2ax+a2+2by-b2=9
which passes through 0,b
i.e. 02+a2+2b2-b2=9

⇒a2+b2=9

Topic: Mathematics, Circle, JEE Main
If the circles of same radius $a$ and centers at $(2,3)$ and $(5,6)$ cut orthogonally, then $a=$

Options:
A $1$
B $2$
C $3$
D $4$

The correct answer is:
$3$

Solution:
$\left(C_1 C_2\right)^2=r_1^2+r_2^2 \Rightarrow 2 a^2=18 \Rightarrow a=3$

Topic: Mathematics, Circle, JEE Main
If the circles (x+1)2+(y+2)2=r2 and x2+y2-4x-4y+4=0 intersect at exactly two distinct points, then

Options:
A 5<r<9
B 0<r<7
C 3<r<7
D 12<r<7

The correct answer is:
3<r<7

Solution:
Given,

S1≡x+12+y+22=r2 and S2≡x2+y2-4x-4y+4=0 or S2≡x-22+y-22=22

Now, if two circles intersect at two distinct points then r1-r2<C1C2<r1+r2.

⇒r-2<32+42<r+2

⇒r-2<5<r+2

⇒r-2<5 and r+2>5

⇒-5<r-2<5 and r>3

⇒-3<r<7 and r>3

⇒3<r<7

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If the circles $x^2+y^2+2 a x+c y+a=0$ and $x^2+y^2-3 a x+d y-1=0$ intersect in two distinct points $P$ and $Q$ then the line $5 x+$ by $-a=0$ passes through $P$ and $Q$ for

Options:
A
exactly one value of a

B
no value of a

C
infinitely many values of a

D
exactly two values of a

The correct answer is:
no value of a

Solution:
$$
\begin{aligned}
& S_1=x^2+y^2+2 a x+c y+a=0 \\
& S_2=x^2+y^2-3 a x+d y-1=0
\end{aligned}
$$
Equation of radical axis of $S_1$ and $S_2$
$$
\begin{aligned}
& S_1-S_2=0 \\
& \Rightarrow 5 a x+(c-d) y+a+1=0
\end{aligned}
$$
Given that $5 \mathrm{x}+\mathrm{by}-\mathrm{a}=0$ passes through $\mathrm{P}$ and $\mathrm{Q}$
$$
\begin{aligned}
& \Rightarrow \frac{a}{1}=\frac{c-d}{b}=\frac{a+1}{-a} \\
& \Rightarrow a+1=-a^2 \\
& a^2+a+1=0
\end{aligned}
$$
No real value of a.

Topic: Mathematics, Circle, JEE Main, JEE Main 2005
If the circles $x^{2}+y^{2}+2 x+2 k y+6=0$, $x^{2}+y^{2}+2 k y+k=0$ intersect orthogonally, then $k$ is

Options:
A 2 or $-3 / 2$
B $-2$ or $-3 / 2$
C 2 or $3 / 2$
D $-2$ or $3 / 2$

The correct answer is:
2 or $-3 / 2$

Solution:
Since, circles cut orthogonally, if
$$
\begin{aligned}
& & 2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime} \\
\Rightarrow & & \quad 0+2 k^{2}=6+k \\
\Rightarrow & & 2 k^{2}-k-6=0 \\
\Rightarrow & &(2 k+3)(k-2)=0
\end{aligned}
$$
$\Rightarrow \quad k=2$ or $-3 / 2$

Topic: Mathematics, Circle, JEE Main
If the circles x2+y2-10x+16y+89-r2=0 and x2+y2+6x-14y+42=0 have common points, then the number of possible integral values of r is equal to

Options:
A 13
B 14
C 15
D 18

The correct answer is:
18

Solution:
Centre & radius of x2+y2-10x+16y+89-r2=0 are 5,-8 & r respectively
Centre & radius of x2+y2+6x-14y+42=0 are -3,7 & 4 respectively
Distance between the centres of circles is 82+152=17
â17-4â¤râ¤17+4
â13â¤râ¤21
ârâÂ± 21,Â±20,Â±19,Â±18,Â±17,Â±16,Â±15,Â±14,Â±13
Hence, the number of possible values of r is equal to 18

Topic: Mathematics, Circle, JEE Main
If the circles x2+y2-16x-20y+164=r2  and (x-4)2+y-72=36 intersect at two distinct points, then:

Options:
A r>11
B 0<r<1
C 1<r<11
D r=11

The correct answer is:
1<r<11

Solution:
As we know, if two circles intersect each other, then
r1-r2<C1C2<r1+r2 ........(i)
Now for the first circle C18, 10 and r1=r
For the second circle C24, 7 and r2=6
From (i)
r-6<5<r+6

⇒r-6<5 ...ii & 5<r+6 ...iii r-6=r-6, r≥66-r, r<6

from iii r>-1 ...iv

from ii

when -1<r<6 then 6-r<5⇒1<r ...v

when r≥6⇒r-6<5⇒r<11 ...vi

from iv, v, vi we get

⇒ r∈1, 11

Topic: Mathematics, Circle, JEE Main
If the circles x2+y2-16x-20y+164=r2  and (x-4)2+y-72=36 intersect at two distinct points, then:

Options:
A r>11
B 0<r<1
C 1<r<11
D r=11

The correct answer is:
1<r<11

Solution:
As we know, if two circles intersect each other, then
r1-r2<C1C2<r1+r2 ........(i)
Now for the first circle C18, 10 and r1=r
For the second circle C24, 7 and r2=6
From (i)
r-6<5<r+6

⇒r-6<5 ...ii & 5<r+6 ...iii r-6=r-6, r≥66-r, r<6

from iii r>-1 ...iv

from ii

when -1<r<6 then 6-r<5⇒1<r ...v

when r≥6⇒r-6<5⇒r<11 ...vi

from iv, v, vi we get

⇒ r∈1, 11

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If the circles $x^2+y^2=a^2$ and $x^2+y^2-2 g x+g^2-b^2=0$ touch each other externally, then

Options:
A $g=a b$
B $g^2=a^2+b^2$
C $g^2=a b$
D $g=a+b$

The correct answer is:
$g=a+b$

Solution:
According to the condition, $\sqrt{(g-0)^2+(0-0)^2}=a+b \Rightarrow g=a+b$

Topic: Mathematics, Circle, JEE Main
If the circles x-32+y-44=16 and x-72+y-72=9 intersect at points A and B, then the area (in sq. units) of the quadrilateral C1AC2B is equal to (where, C1 and C2 are centres of the given circles)

Options:
A 6
B 12
C 18
D 24

The correct answer is:
12

Solution:
So, area of quadrilateral =2 area âAC1C2

=21234=12 square units

Topic: Mathematics, Circle, JEE Main