Let O be the origin and OP and OQ 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 OPQ passes through the point ( 1 2 ) , then a value of is [2023]

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Solution

Q.
Let O be the origin and OP and OQ 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 OPQ passes through the point $(α, \frac{1}{2}),$ then a value of $α$ is [2023]

1 $\frac{5}{2}$

2 $\frac{3}{2}$

3 $- \frac{1}{2}$

4 $1$

Ans.
(1)

PQ is the chord of contact of the tangents from the origin to the circle, $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ ...(i)

Equation of $P Q$ is, $3 x - 2 y - 8 = 0$ ...(ii)

Equation of circle passing through the intersection of (i) and (ii) is, $x^{2} + y^{2} - 6 x + 4 y + 8 + λ (3 x - 2 y - 8) = 0$ ...(iii)

If this represents the circumcircle of the triangle $O P Q$ , it passes through $O (0, 0)$ .

So, $λ = 1$ and (iii) becomes

$x^{2} + y^{2} - 3 x + 2 y = 0$ ...(iv)

Given, $(α, \frac{1}{2})$ passes through (iv)

$∴$ $α^{2} + \frac{1}{4} - 3 α + 1 = 0$ $\Rightarrow 4 α^{2} - 12 α + 5 = 0$

$\Rightarrow (2 α - 1) (2 α - 5) = 0$ $\Rightarrow α = \frac{1}{2} or \frac{5}{2}$

Hence, the value of $α$ is $\frac{5}{2}$ .

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