Orthogonal circles What is the equation of the circle that is orthogonal to the circles  $x^2 + y^2 - 8x +5 =0$ and $x^2 + y^2 +6x +5 = 0$ and passes through the point $(3,4)$? I've spent hours trying to figure this out - help please

you have $$C_1:(x-4)^2+y^2=11,\quad C_2:(x+3)^2+y^2= 4$$  we will look for a point $(a,0)$ on the $x$-axis so that the tangents to the circle have the same length. 
we have $$(4-a)^2 -11=(a+3)^2-4 \to-8a+5=6a+5\to a = 0.$$ 
we will use the fact that if a circle is orthogonal to both $C_1, C_2,$  then the $x$-cdt of the center is $0.$ 
let the the center of $C$ that is orthogonal to both $C_1, C_2$ and goes through $(3,4)$ be $(0,b)$.
$$x^2 + (y-b)^2 = b^2 + 9-4 \to x^2+y^2 - 2by=5$$ making this go through $(3,4)$ requires $$3^2 + 4^2-8b=5\to b= \frac 52 $$ andn the circle is $$x^2 + y^2-5y=5. $$
