JEE MAIN Circle Previous Year Questions (PYQs) – Page 1 of 8
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If the angle of intersection at a point where two circles with radii 5cm and 12cm intersect is $90^\circ$, then the length (in cm) of their common chord is:
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Solution
$AB = \dfrac{2 \cdot r_1 \cdot r_2}{\sqrt{r_1^2 + r_2^2}}$
$ = \dfrac{2 \times 5 \times 12}{\sqrt{25 + 144}} = \dfrac{120}{\sqrt{169}}$
$ = \dfrac{120}{13}$
$\boxed{AB = \dfrac{120}{13} \text{ cm}}$
Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC divides the arc AC such that $\dfrac{\text{length of arc }AB}{\text{length of arc }BC}=\dfrac{1}{5}$, and $\overrightarrow{OC}=\alpha\,\overrightarrow{OA}+\beta\,\overrightarrow{OB}$, then $\alpha+\sqrt{2}\,(\sqrt{3}-1)\,\beta$ is equal to:
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Solution
Let radius $=1$ and take $O$ as origin
Arc $AC$ subtends $\frac{\pi}{2}$ ⇒ take $A(1,0),\ C(0,1)$
Given $\frac{AB}{BC}= \frac{1}{5}$ ⇒ $\angle AOB=\frac{\pi}{12}$
So $B=\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
$\overrightarrow{OC}=\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}$
$(0,1)=\alpha(1,0)+\beta\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
Comparing components:
$\alpha+\beta\cos\frac{\pi}{12}=0$
$\beta\sin\frac{\pi}{12}=1$
$\beta=\frac{1}{\sin\frac{\pi}{12}}=\frac{1}{\frac{\sqrt6-\sqrt2}{4}}=\sqrt6+\sqrt2$
$\alpha=-\beta\cos\frac{\pi}{12}=-(\sqrt6+\sqrt2)\cdot\frac{\sqrt6+\sqrt2}{4}=-2-\sqrt3$
Now,
$\alpha+\sqrt2(\sqrt3-1)\beta$
$=(-2-\sqrt3)+\sqrt2(\sqrt3-1)(\sqrt6+\sqrt2)$
$\sqrt2(\sqrt6+\sqrt2)=2\sqrt3+2$
So expression:
$=(-2-\sqrt3)+(2\sqrt3+2)(\sqrt3-1)$
$=(-2-\sqrt3)+(6-2\sqrt3+2\sqrt3-2)$
$=(-2-\sqrt3)+4=2-\sqrt3$
$\boxed{2-\sqrt3}$
Arc $AC$ subtends $\frac{\pi}{2}$ ⇒ take $A(1,0),\ C(0,1)$
Given $\frac{AB}{BC}= \frac{1}{5}$ ⇒ $\angle AOB=\frac{\pi}{12}$
So $B=\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
$\overrightarrow{OC}=\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}$
$(0,1)=\alpha(1,0)+\beta\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
Comparing components:
$\alpha+\beta\cos\frac{\pi}{12}=0$
$\beta\sin\frac{\pi}{12}=1$
$\beta=\frac{1}{\sin\frac{\pi}{12}}=\frac{1}{\frac{\sqrt6-\sqrt2}{4}}=\sqrt6+\sqrt2$
$\alpha=-\beta\cos\frac{\pi}{12}=-(\sqrt6+\sqrt2)\cdot\frac{\sqrt6+\sqrt2}{4}=-2-\sqrt3$
Now,
$\alpha+\sqrt2(\sqrt3-1)\beta$
$=(-2-\sqrt3)+\sqrt2(\sqrt3-1)(\sqrt6+\sqrt2)$
$\sqrt2(\sqrt6+\sqrt2)=2\sqrt3+2$
So expression:
$=(-2-\sqrt3)+(2\sqrt3+2)(\sqrt3-1)$
$=(-2-\sqrt3)+(6-2\sqrt3+2\sqrt3-2)$
$=(-2-\sqrt3)+4=2-\sqrt3$
$\boxed{2-\sqrt3}$
Let $P$ and $Q$ be any points on the curves $(x-1)^{2}+(y+1)^{2}=1$ and $y=x^{2}$, respectively.
The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval:
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Solution
Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2}-4x-6=0$
and the ordinates of $P$ and $Q$ be the roots of $y^{2}+2y-7=0$.
If $PQ$ is a diameter of the circle $x^{2}+y^{2}+2ax+2by+c=0$, then the value of $(a+b-c)$ is _________.
(A)
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Solution
If $P$ is a point on the circle $x^2 + y^2 = 4$, $Q$ is a point on the straight line $5x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of abscissa of all such point $P$ is ______.
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Solution
Let $P(2\cos\theta,; 2\sin\theta)$
$Q(\alpha,; -5\alpha - 2)$
Mid point of $PQ$ lies on $x - y + 1 = 0$
$\Rightarrow \frac{2\cos\theta + \alpha}{2} - \frac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$
$\Rightarrow 2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$
$\Rightarrow \cos\theta - \sin\theta + 3\alpha + 2 = 0 \quad ...(1)$
$\because$ slope of $PQ = -1$
$\Rightarrow \frac{2\sin\theta + 5\alpha + 2}{2\cos\theta - \alpha} = -1$
$\Rightarrow 2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$
$\Rightarrow \sin\theta + \cos\theta + 2\alpha + 1 = 0 \quad ...(2)$
Eliminate $\alpha$ from (1) and (2)
$\Rightarrow \cos\theta + 5\sin\theta = 1,; \theta \in [0, 2\pi]$
$\Rightarrow 5\times 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin^2\frac{\theta}{2}$
$\therefore \sin\frac{\theta}{2} = 0 \Rightarrow \cos\theta = 1$
or
$\sin\frac{\theta}{2} = \frac{5}{2}\cos\frac{\theta}{2} \Rightarrow \cos\theta = -\frac{12}{13}$
Sum of all possible values of abscissa of point $P$ is
$= 2\cdot 1 + 2\left(-\frac{12}{13}\right) = \frac{2}{13}$
$\therefore 13$ times sum of all values of abscissa of point $P$ is $2$
$\therefore 13$ times sum of all values of abscissa of point $P$ is $2$
Let $C$ be a circle with radius $\sqrt{10}$ units and centre at the origin.
Let the line $x+y=2$ intersect the circle $C$ at the points $P$ and $Q$.
Let $MN$ be a chord of $C$ of length $2$ units and slope $-1$.
Then, the distance (in units) between the chord $PQ$ and the chord $MN$ is:
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Solution
If the area of an equilateral triangle inscribed in the circle $x^{2}+y^{2}+10x+12y+c=0$ is $27\sqrt{3}$ sq units, then $c$ is equal to :
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Solution
Let $A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix}$.
Let $\alpha, \beta \in \mathbb{R}$ be such that $\alpha A^2 + \beta A = 2I$.
Then $\alpha + \beta$ is equal to :
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Solution
A circle touching the $x$-axis at $(3,0)$ and making an intercept of length $8$ on the $y$-axis passes through the point:
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Solution
Let $C_1$ be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let $C_2$ be the circle with centre $(1,3)$ that touches $\mathrm{C}_1$ externally at the point $(\alpha, \beta)$. If $(\beta-\alpha)^2=\frac{m}{n}$ , $\operatorname{gcd}(m, n)=1$, then $m+n$ is equal to
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Solution
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