Equation of a circle cutting another given circle orthogonally

Question

What is the equation of the circle which touches the line $x+y=5$ at $(-2,7)$ and cut the circle $$x^2+y^2+4x-6y+9=0$$ orthogonally?

I tried to denote the center of circle as $(h,k)$ and radius as $r$. Orthonogallity implies \begin{align*} (h+2)^2+(k-3)^2&=r^2+4,\\ (h+2)^2+(k-7)^2&=r^2. \end{align*} Solving equation gives $k=\dfrac{11}{2}$. How to find $h,r$?

Answer 1

The equation of tangent to a circle is given by $T=0$ Let the equation of required circle be $x^2+y^2+2gx+2fy+c=0$ and the tangent to that circle is at point $(-2,7)$. Hence using above information the equation of tangent at that point is given by $$(g-2)x+(f+7)y-2g+7f+c=0$$ But the equation of tangent is given to be $x+y-5=0$ Hence we get $$\frac {g-2}{1}= \frac {f+7}{1}=\frac {2g-7f-c}{5}$$ Using this information we get $g=f+9$ and $c=-10f-17$ Using the condition of orthogonality you already got $f=\frac {-11}{2}$

Now you just have to substitute the values to get the equation of required circle.

Answer 2

HINT....In addition to what you have done, you need to use the fact that since the circle touches the line at $(-2,7)$ then the gradient of the line from the centre $(h,k)$ to this point is perpendicular to the line. This gives you another equation for $h,k$.