Equation of circumcircle

Question

From a point $(a,b)$ two tangents $\overset{\leftrightarrow}{PQ}$ and $\overset{\leftrightarrow}{PR}$ are drawn to a circle $x^2 + y^2 - a^2=0$ Find the equation of the circumcircle of $\triangle PQR$.

My attempt: The circumcircle and the given circle have a common chord $\overline{QR}$.

Apart from this I could not convert this into useful information.

Answer 1

$$x^2+y^2-a^2=0\ \ \ \ (0)$$ Using The equation of a pair of tangents to a circle from a point. tangents-to-a-circle-from-a-point,

the equation of the pair of tangents, $$(a^2+b^2-a^2)(x^2+y^2-a^2)=(ax+by-a^2)^2\ \ \ \ (1)$$

Now the equation of any conic passing through the intersection of $(0),(1)$ will be $$(a^2+b^2-a^2)(x^2+y^2-a^2)-(ax+by-a^2)^2+K(x^2+y^2-a^2)=0\ \ \ \ (2)$$ where $K$ is an arbitrary constant.

Now $(2)$ has to pass through $(a,b)$

Answer 2

HINT:

The equation of tangent at $P(a\cos2t,a\sin2t)$ is $$x\cos2t+y\sin2t=a$$

Now if this passes through $(a,b),$

$$a\cos2t+b\sin2t=a\iff2b\sin t\cos t=a(2\sin^2t)\implies$$

either $\sin t=0\implies P_1(a,0)$

or $\tan t=\dfrac ba\implies P_2\left(a\cdot\dfrac{a^2-b^2}{a^2+b^2},\dfrac{2a^2b}{a^2+b^2}\right)$

We already have $$Q(a,b)$$

Now use this

Answer 3

Let $O$ be the centre of the circle. Since $\angle QPR$ and $\angle QOR$ are supplementary, we have $P,Q,R, O$ are concylic.

That means circumcircle of $\triangle PQR$ passes through $O$, and further since $\angle PQO = 90^{\circ}$, $OP$ is a diameter of the circle.

Hence the midpoint of the $OP$ i.e. $\left(a, \frac{b}{2} \right)$ is the centre of the circle and radius $= \frac{b}{2}$, which is sufficient to write the equation of the circle