Question

Find the equation of the director circle of the circle ${(x-h)}^2+{(y-k)}^2=a^2$

• A. ${(x-h)}^2+{(y-k)}^2=\sqrt{2}{a}^2$
• B. ${(x-h)}^2+{(y-k)}^2=2a^2$
• C. ${(x+h)}^2+{(y+k)}^2=2a^2$
• D. ${(x+h)}^2+{(y+k)}^2=\sqrt{2}{a}^2$

Answer 1

Answer: (B)

${(x-h)}^2+{(y-k)}^2=2a^2$

The director circle is the locus of all those points from which, if a pair of tangents are drawn to the given circle, then they are always mutually perpendicular.

Consider the point $P(l,m)$.

Since the tangents are mutually perpendicular $\angle P={90}^0$

Therefore, $\angle \frac{P}{2}={45}^0$.

Let the point of contact of the tangent and the circle be $A$ and $B$.

And let the center of the circle be $C$.

Then in triangle $PAC$,

$PA=CA$ since the triangles are isosceles right angled triangle.

Therefore, $PC=\sqrt{2}a$

Hence, the center is the origin, hence

$P{C}^2=2a^2$

$(l-h{)}^2+(m-k{)}^2=2a^2$

Replacing $(h,k)$ by $(x,y)$, we get

$(x-h{)}^2+(y-k{)}^2=2a^2$