Analytic Geometry Question: Computing the equation of a circle given 2 points and center I want to derive the equation of a circle passing through the intersection of the two circles. My method gave me a nonsensical solution. I would love if someone could tell me how my method went wrong, although since my solution is kinda long (but it's mostly algebra) I would welcome an alternative solution. 
$$
x^2+y^2=100\\
(x-11)^2+(y-4)^2=9
$$
and the centered at $(22,8)$. 
The way I went about solving this was first starting with the family of curves $F(x,y,h,k)=0$ which go through the point of intersection
$$
h(x^2+y^2-100)+k(x-11)^2+(y-4)^2-9)=0\\
\Rightarrow (h+k)(x^2+y^2)-22kx-8ky+121k+16k-9k-100h=0\\
\Rightarrow (h+k)(x^2+y^2)-22kx-8ky+128k-100h=0\\
\Rightarrow x^2+y^2-\frac{22k}{h+k}x-\frac{8k}{h+k}y+\frac{128k}{h+k}-\frac{100h}{h+k}=0\\
\Rightarrow(x-\frac{11k}{h+k})^2+(y-\frac{4k}{h+k})^2
+\frac{128k+100h}{h+k}=\frac{121k^2+16k^2}{(h+k)^2}=\frac{137k^2}{(h+k)^2}\\
\Rightarrow (x-\frac{11k}{h+k})^2+(y-\frac{4k}{h+k})^2=\frac{137k^2}{(h+k)^2}-\frac{128k+100h}{h+k}
$$
Then imposing the restriction that this circle is centered at $(22,8)$ we have the system
$$
\frac{11k}{h+k}=22\\
\frac{4k}{h+k}=8\Rightarrow 2h=-k
$$
Which yields a potential solution of $(h,k)=(1,-2)$ and a radius of $392$.
edit: I fixed an algebraic mistake at the end, would still be curious to hear alternative solutions!
 A: First of all, it would be slightly easier if we write the resultant circle as:-
$((x-11)^2+(y-4)^2-9) + j(x^2+y^2-100) = 0$
because only one variable is involved.
Secondly, the painful completing square process can be avoided. 
Therefore the equation of the required circle is $x^2 + y^2 + 2(\dfrac {-11}{1 + j})x + 2(\dfrac {-4}{1 + j})y + \dfrac {121 + 16 – 9 -100j}{1 + j} = 0$
Correction:-
From $\dfrac {-11}{1 + j} = -22$, we get $j = -0.5 (= h/k)$
Then, the radius $= \sqrt {22^2 + 8^2 -  356}$
A: Less elegant than Mick's answer.
Considering $$x^2+y^2=100 \tag 1$$ $$(x-11)^2+(y-4)^2=9 \tag 2$$ Subtract $(1)$ from $(2)$ to get $$-22 x-8 y+228=0\implies y=\frac{1}{4} (114-11 x)\tag 3$$ Replace in $(1)$ to get, after simplification $$137 x^2-2508 x+11396=0\tag 4$$ One of the solutions corresponds to $$x=\frac{2}{137} \left(627-16 \sqrt{11}\right)\implies y=\frac{8}{137} \left(57+11 \sqrt{11}\right)$$ So, $$(x-22)^2+(y-8)^2=r^2$$ Replacing by the above values of $x,y$ leads to $r^2=192$.
I then suppose that I made some mistakes somewhere. However, a plot of the three circles seems to confirm this answer.
