JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Circle

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 point 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

Answer: (B)

Solution

The circle

x^{2} + y^{2} - 6 x + 4 y + 8 = 0

has centre

C (3, - 2)

. As

O P

and

O Q

are tangents from

O

∠ O P C = ∠ O QC = 9 0^{\circ}

, so

P

and

Q

lie on the circle with

O C

as diameter; this circle, passing through

O, P, Q

, is the circumcircle of

△ O P Q

. In diameter form,

(x - 0) (x - 3) + (y - 0) (y + 2) = 0 \Rightarrow x^{2} + y^{2} - 3 x + 2 y = 0.

Putting

(α, \frac{1}{2})

on it:

α^{2} - 3 α + \frac{1}{4} + 1 = 0

, i.e.

α^{2} - 3 α + \frac{5}{4} = 0

, giving

α = \frac{3 \pm 2}{2}

. The value matching the given choices is

α = \frac{5}{2}

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option