Let be the origin and and be the tangents to the circle at the points and on it. If the circumcircle of the triangle passes through the point , then a value of is:

Explanation: Centre Equation of circumcircle is Since is on the circle So

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $O$ be the origin and $O P$ and $O Q$ be the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $O P Q$ passes through the point $(α, \frac{1}{2})$ , then a value of $α$ is:

Choose the correct answer:

A.
$- \frac{1}{2}$
B.
$\frac{5}{2}$
(Correct Answer)
C.
1
D.
$\frac{3}{2}$

Correct Answer:

$\frac{5}{2}$

Explanation

Centre $(3, - 2)$
Equation of circumcircle is $x (x - 3) + y (y + 2) = 0$
$\Rightarrow x^{2} - 3 x + y^{2} + 2 y = 0$

Since $(α, \frac{1}{2})$ is on the circle
So $α^{2} - 3 α + \frac{1}{4} + 1 = 0$
$\Rightarrow 4 α^{2} - 12 α + 5 = 0$

$\Rightarrow α = \frac{12 \pm 144 - 80}{8}$
$= \frac{12 \pm 64}{8} = \frac{12 \pm 8}{8}$
$α = \frac{20}{8}, \frac{4}{8} = \frac{5}{2}, \frac{1}{2}$

Explanation

Centre $(3, - 2)$
Equation of circumcircle is $x (x - 3) + y (y + 2) = 0$
$\Rightarrow x^{2} - 3 x + y^{2} + 2 y = 0$

Since $(α, \frac{1}{2})$ is on the circle
So $α^{2} - 3 α + \frac{1}{4} + 1 = 0$
$\Rightarrow 4 α^{2} - 12 α + 5 = 0$

$\Rightarrow α = \frac{12 \pm 144 - 80}{8}$
$= \frac{12 \pm 64}{8} = \frac{12 \pm 8}{8}$
$α = \frac{20}{8}, \frac{4}{8} = \frac{5}{2}, \frac{1}{2}$

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