Find the center of circle $P$ and $Q$

Question

Given that circle $P : Center(a,b)$ has a radius of $r_1$, circle $ Q : Center(c,d)$ has a radius of $r_2$. Circle $P$ and circle $Q$ intersects at point $A = (3,8)$ and $B = (3,-4)$ when $r_2=\frac{2}3r_1$.

If $$a,b,c,d \subseteq \mathbb Z$$ when $$30 \lt 3c-2a \lt 40$$

Q : Find the center of circle $P$ and $Q$.

I have determined the pythagoras triangle between two circles; $$a_1=6,b_1=|3-a|,c_1=r_1$$ and $$a_2=6,b_2=|3-c|,c_2=\frac{2}3*r_1$$ Which using one of the many circle properties, I come up with $$|3-a|=|3-c|$$ which i couldn’t progress further.

Help me solve the problem anew or continue with my solution.

Answer 1

I will not hesitate to put that into mathjax though. For reference only.