Your first problem is that you have the same parameters, h, k, and r, in both equations! Remember that (h, k) is the center of the circle. Since these circles do not have the same center those must be different. You might show that by using "\(\displaystyle h_1\), \(\displaystyle k_1\)", "\(\displaystyle r_1\)" for one circle, "\(\displaystyle h_2\), \(\displaystyle k_2\)", "\(\displaystyle r_2\)" for the other.
So your equations are
\(\displaystyle (x- h_1)^2+ (y- k_1)^2= r_1^2\) and
\(\displaystyle (x- h_2)^2+ (y- k_2)^2= r_2^2\)
The other problem is that those are NOT linear equations and matrices work for linear equations, not non-linear equations.
You could try this:
Multiply out those squares:
\(\displaystyle x^2- 2h_1x+ h_1^2+ y^2- 2k_1y+ k_1^2= r_1^2\)
\(\displaystyle x^2- 2h_2x+ h_2^2+ y^2- 2k_2y+ k_2^2= r_2^2\)
Subtract the second equation from the first, eliminating the squares:
\(\displaystyle 2(h_2- h_1)x+ (h_1^2- h_2^2)+ 2(k_2- k_1)y+ (k_1^2- k_2^2)= r_1^2- r_2^2\)
We have the linear equation
\(\displaystyle 2(h_2- h_1)x+ 2(k_2- k_1)y= r_1^2- r_2^2+h_1^2- h_2^2+ k_1^2- k_2^2\).
But that is only one equation- we need two linear equation to use matrix methods.
It appears, from you picture that the two circles have the same radii, \(\displaystyle r_1= r_2\). If that is true then the x coordinate of the two points of intersection are half way between \(\displaystyle h_1\) and \(\displaystyle h_2\): \(\displaystyle x= \frac{h_1+ h_2}{2}\). Then we can solve either \(\displaystyle (\frac{h_1+ h_2}{2}- h_1)^2+ (y- k_1)^2= \frac{h_2- h_1)^2}{4}+ (y- k_1)^2= r_1^2\) or \(\displaystyle (\frac{h_1+ h_2}{2}- h_2)^2+ (y- k_1)^2= \frac{h_2- h_1)^2}{4}+ (y- k_2)^2= r_2^2\) for the two y values.
