Question
Tangents drawn from the point $$P\left( {1,\,8} \right)$$ to the circle $${x^2} + {y^2} - 6x - 4y - 11 = 0$$ touch the circle at the points $$A$$ and $$B.$$ The equation of the circumcircle of the triangle $$PAB$$ is-
A.
$${x^2} + {y^2} + 4x - 6y + 19 = 0$$
B.
$${x^2} + {y^2} - 4x - 10y + 19 = 0$$
C.
$${x^2} + {y^2} - 2x + 6y - 29 = 0$$
D.
$${x^2} + {y^2} - 6x - 4y + 19 = 0$$
Answer :
$${x^2} + {y^2} - 4x - 10y + 19 = 0$$
Solution :
Tangents $$PA$$ and $$PB$$ are drawn from the point $$P\left( {1,\,3} \right)$$ to circle $${x^2} + {y^2} - 6x - 4y - 11 = 0$$ with centre $$C\left( {3,\,2} \right)$$
Clearly the circumcircle of $$\Delta PAB$$ will pass through $$C$$ and as $$\angle A = {90^ \circ },$$
$$PC$$ must be a diameter of the circle.
$$\therefore $$ Equation of required circle is
$$\eqalign{ & \left( {x - 1} \right)\left( {x - 3} \right) + \left( {y - 8} \right)\left( {y - 2} \right) = 0 \cr & \Rightarrow {x^2} + {y^2} - 4x - 10y + 19 = 0 \cr} $$
Tangents $$PA$$ and $$PB$$ are drawn from the point $$P\left( {1,\,3} \right)$$ to circle $${x^2} + {y^2} - 6x - 4y - 11 = 0$$ with centre $$C\left( {3,\,2} \right)$$
Clearly the circumcircle of $$\Delta PAB$$ will pass through $$C$$ and as $$\angle A = {90^ \circ },$$
$$PC$$ must be a diameter of the circle.
$$\therefore $$ Equation of required circle is
$$\eqalign{ & \left( {x - 1} \right)\left( {x - 3} \right) + \left( {y - 8} \right)\left( {y - 2} \right) = 0 \cr & \Rightarrow {x^2} + {y^2} - 4x - 10y + 19 = 0 \cr} $$