Question

If the diameter of the circumcircle of $\Delta TPQ$, where PQ is the chord of contact corresponding to the point T with respect to the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-11=0$, is 12 units, then the minimum 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

Hint:
We first try to form the circumcircle of the $\Delta TPQ$, where PQ is the chord of contact corresponding to the point T with respect to the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-11=0$. The circumcircle goes through the centre of the given circle. We form the intersecting line of the point T and the centre C. From there we find the minimum distance of T from the director circle of the given circle.

Complete step by step answer:
The radius of the circumcircle of $\Delta TPQ$, where PQ is the chord of contact corresponding to the point T with respect to the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-11=0$, is 12 units.
We need to find the minimum distance of T from the director circle of the given circle ${{x}^{2}}+{{y}^{2}}-2x+4y-11=0$.
The given circle can be expressed in its general form as ${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{4}^{2}}$. Equating with general equation of circle ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$, we get the centre as $C\equiv \left( 1,-2 \right)$ and the radius as 4 units.
We know that the director circle is a concentric circle whose radius is $\sqrt{2}$ times the radius of the given circle.
So, the director circle is ${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{\left( 4\sqrt{2} \right)}^{2}}$ which gives ${{x}^{2}}+{{y}^{2}}-2x+4y=27$.
We draw the diagram for the given equations and get

The radius of the circumcircle is 12 units. PQ is the chord of contact with respect to the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-11=0$. TS be the minimum distance of T from the director circle of the given circle ${{x}^{2}}+{{y}^{2}}-2x+4y-11=0$.
The segment joining the centre of the given circle as well as the director circle as C with the vertices T of the $\Delta TPQ$ becomes the diameter of the circumcircle.
So, $TS=CT-CS=12-4\sqrt{2}$. The correct option is D

Note:
 We need to remember that the circumcircle was drawn in such a way that it touches the center C. it was given that PQ is the chord of contact corresponding to the point T with respect to the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-11=0$. Otherwise, we couldn’t have expressed the joining segment as a single line.