Let O be the origin and OP and OQ be the tangents to the circle x²+y²-6x+4y+8=0 at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point ( α ,1/2 ), then a value of α is :

Solution

Maths sir · 4 steps

Each tangent is perpendicular to its radius, so O, P, Q and centre C lie on the circle with diameter OC.

1. Centre of the circle. Compare with x²+y²+2gx+2fy+c=0. The centre is C(3,-2) and the radius is √(9+4-8)=√5. Since c = 8 > 0, the origin lies outside the circle, so two tangents from O exist.
2. Right angles. A tangent is perpendicular to the radius at its point of contact, so ∠ OPC = ∠ OQC = 90°. So OC subtends a right angle at P and at Q, which puts P and Q on the circle with OC as diameter. That circle also passes through O, so it is the circumcircle of triangle OPQ.

   (x-0)(x-3)+(y-0)(y+2)=0 ⇒ x²+y²-3x+2y=0

3. Put in the point. Substitute (α, 1/2): α²+1/4-3α+1=0, so α²-3α+5/4=0. Then α=(3±√(9-5))/2=(3± 2)/2, which gives α=5/2 or α=1/2.
4. So. Of the two values, only 5/2 appears among the options. The answer is (c).

The idea to keep. Whenever tangents are drawn from an external point T, the two points of contact, T and the centre are concyclic. Their circle has the segment from T to the centre as diameter.

Answer from the NTA final answer key. Solution by PrepWiser.