Question

Tangents drawn from $P(1,8)$ to the circle $x^2+y^2-6x-4y$ $-11=0$ touches the circle at the points A and B, respectively. The radius of the circle which passes through the points of intersection of circles $x^2+y^2-2x+6y-6=0$ and $x^2+y^2-2x-6y+6=0$ and intersects the circumcircle of the $\Delta PAB$ orthogonally is equal to

• A. $\frac{\sqrt{73}}{4}$
• B. $\frac{\sqrt{71}}{2}$
• C. 3
• D. 2

Answer 1

Answer: (A)

Equation of circle circumscribing $\Delta PAB$ is:
$(x-1)(x-3)+(y-8)(y-2)=0$
$\Rightarrow x^2+y^2-4x-10y+19=0$
Equation of circle passing through points of intersection of circles $x^2+y^2-2x-6y+6=0$ and $x^2+$ $y^2-2x-6y+6=0$ is given by
$\left(x^2+y^2-2x-6y+6\right)+\lambda \left(x^2+y^2-2x-6y-6\right)=0$
$\Rightarrow x^2+y^2+\frac{(2\lambda -2)}{\lambda +1}x-6y+6$
As circle (ii) is orthogonal to circle (i), we have
$-2\left(\frac{2\lambda -2}{\lambda +1}\right)-5(-6)=19+6$
$\Rightarrow 4\lambda -4=5\lambda +5$
$\Rightarrow \lambda =-9$
Hence, required equation fo circle is:
$x^2+y^2+\frac{5}{2}x-6y+6=0$
$\therefore$ Radius of circle $=\sqrt{\frac{25}{16}+9-6}=\frac{\sqrt{73}}{4}$