Question

Tangents $PQ,PR$ are drawn to the circle $x^2+y^2=a^2$ from a given point $P$. Find the equation of the circum-circle of the triangle $PQR$.

Answer 1

Let $P$ be $(x_1,y_1)$ so that $QR$ is chord of contact whose equation is $x{x}_1+y{y}_1-a^2=0$. The circle $PQR$ thus passes through the intersection of given circle and chord $QR$ and hence by $S-\lambda P=0$ its equation is
$(x^2+y^2-a^2)-\lambda (x{x}_1+y{y}_1-a^2)=0.....\left(1\right)$
As it passes through $P(x_1,y_1)$
$\therefore (x_1^2+y_1^2-a^2)-\lambda (x_1^2+y_1^2-a^2)=0.\therefore \lambda =1$
Putting the value of $\lambda$ in $\left(1\right)$ the required circle is $x^2+y^2-x{x}_1-y{y}_1=0$ which clearly passes through the centre $(0,0)$ of the given circle.
Another form : The equation of the circle can also be written as $(x-0)(x-x_1)+(y-0)(y-y_1)=0$
i.e., it is on $CP$ as diameter.