Question

If the radius of the circumcircle of the triangle TPQ where PQ is chord of contact corresponding to point T with respect to circle $x^2+y^2-2x+4y-11=0$ is $6$ units then distance of T from the director circle of the given circle is

• A. $6$
• B. $12$
• C. $6\sqrt{2}$
• D. $12$ - $4\sqrt{2}$

Answer 1

The correct option is D $12$ - $4\sqrt{2}$

Radius of circumcircle of triangle TPQ is $6$

$PQ$ is chord of contact corresponding to point $T$ on circle $x^2+y^2-2x+4y=11$

$(x-1{)}^2+(y+2{)}^2=4^2$

Centre$(1,-2)r=4$

Diameter of circumcircle

$D=\frac{a}{\sin A}PQ=2r\cos \theta PQ=8\cos \theta$

Circumradius $=6,2\theta$ is the angle between two tangent

$12=\frac{8\cos \theta }{\sin 2\theta }=\frac{8\cos \theta }{2\sin \theta \cos \theta }\sin \theta =\left(\frac{1}{3}\right)\cos \theta =\frac{2\sqrt{2}}{3}\sin \theta =\frac{OP}{OT}\Rightarrow \frac{1}{3}=\frac{4}{OT}\Rightarrow OT=12cm$

Distance between centre of circle and the boundary of the director circle $=\sqrt{2}a=4\sqrt{2}$

Distance between $T$ and director circle $=12-4\sqrt{2}$