Question

From a point $P$ tangents drawn to the circles $x^2+y^2+x-3=0,3x^2+3y^2-5x+3y=0$ and $4x^2+4y^2+8x+7y+9=0$ are of equal length. Find the equation of the circle through $P$ which touches the line $x+y=5$ at the point $(6,-1)$.

Answer 1

Proceeding as in part (a) the co-ordinates of the radical centre are found to be $P(0,-3)$. Now we have to find a circle which passes through $P(0,-3)$ and touches the line $x+y=5$ at $(6,-1)$.
The required circle by rule $\left(n_1\right)$ is
$(x-6{)}^2+(y+1{)}^2+\lambda (x+y-5)=0$
It passes through $P(0,-3)$.
$\therefore 36+4+\lambda (-8)=0\therefore \lambda =5$
Putting in $\left(1\right)$, the required circle is
$x^2+y^2-7x+7y+12=0$.