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 the 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 $12-4\sqrt{2}$

Given circle $SL(x–1{)}^2+(y–2{)}^2=4^2,c=(-1,2),r=2$

$PQ$ is COC corresponding to $T$

$PT$ and $QT$ are tangents

WKT the circumcircle of $QTP$ passes through $C$ since $CPTQ$ is a cyclic quadrilateral with $\angle CPT=\angle CQT={90}^{\circ }$

Also, $CT$ will be the diameter.

Given radius of the above circle is $6$.

$⟹CT=2r=12$

Now equation of director circle pf $5$ is $(x-1{)}^2+(y–2{)}^2=2\times 4^2=(4\sqrt{2}{)}^2$

Let the line $CT$ meet the director circle at $A⟹CA$ is radius of director circle.

$⟹CA+AT=CT$

$⟹AT=CT-CA$

$AT=12–4\sqrt{2}$

Distance of $T$ from director circle is $12–4\sqrt{2}$