question_answer

The tangent and normal at the point P (18, 12) of the parabola \[{{y}^{2}}=8x\] intersects the x-axis at the points Q and R respectively. The equation of the circle through P, Q and R is given by

A) \[{{x}^{2}}+{{y}^{2}}-4x-396=0\]

B) \[{{x}^{2}}+{{y}^{2}}-4x-360=0\]

C) \[{{x}^{2}}+{{y}^{2}}-4x-346=0\]

D) \[{{x}^{2}}+{{y}^{2}}-4x-300=0\]

Correct Answer: A

Solution :

We know \[PS=QS=SR\] Hence, S is circumventer of \[\Delta PQR\] Equation of required circle \[{{(x-2)}^{2}}+{{(y-0)}^{2}}={{(2-18)}^{2}}+{{(-12)}^{2}}\] \[{{x}^{2}}+{{y}^{2}}-4x-396=0\]