Finding orthogonal circle

Question

How to find the equation of a circle which is orthogonal to both the circles $x^2+y^2=4$ and $x^2+y^2-8x-8y+28=0$?? Me got the equation of radical axis, now i thought the centre of the required circle lies on radical axis ,but how to proceed further?

Answer 1

Any point $\,P\,$ on the radical axis will have a single circle centered at $\,P\,$ which is orthogonal to both given circles. Consider the four tangents from point $\,P\,$ to the two given circles. Since $\,P\,$ is on the radical axis, it has equal power with respect to both circles, thus the squares of the lengths of all four tangents (and therefore the lengths of all four tangents) must be equal. But this means that a circle centered at $\,P\,$ with radius $\,R,\,$ where $\,R^2\,$ is the power of $\,P\,$ with respect to the two circles, must be orthogonal to both circles.

In your case, the radical axis is $\,x+y=4,\,$ so taking any point on that line and computing its power will lead to the equation of an orthogonal circle. Some examples would be:

$$ \begin{align} &P(1,3): \quad && R^2 = 10-4=6 &&&\implies &&& \quad (x-1)^2+(y-3)^2=6\\ &P(6,-2): \quad && R^2 = 40-4=36 &&&\implies &&& \quad (x-6)^2+(y+2)^2=36\\ \end{align} $$

and a general formula for all circles orthogonal to your given circles would be

$$(x-2+\alpha)^2 + (y-2-\alpha)^2 = 2(\alpha^2+2) \quad {\small\text{for any }} \alpha \in \Bbb{R}$$

[Note that if the two given circles were intersecting, $\,P\,$ would have to lie outside the circles.]