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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.

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  • $\begingroup$ Is $I$ meant to be the irrational numbers ($\Bbb R - \Bbb Q$) or the integers ($\Bbb Z$)? $\endgroup$ – Patricio Dec 17 '18 at 8:53
  • $\begingroup$ $$r_1^2=(a-3)^2+(b+4)^2=(a-3)^2+(b-8)^2$$ $\implies b=2,r_1^2=(a-3)^2+(2+4)^2$ As $(3,8)$ is on $P,r_1^2=(3-3)^2+6^2\implies r_1=6$ So, if the equation of $P$ is $$(x-a)^2+(y-2)^2=6^2$$ and $(3,8)$ lies on $P,a=3$ $\endgroup$ – lab bhattacharjee Dec 17 '18 at 8:56
  • $\begingroup$ a subset I is meaninglesx. $\endgroup$ – William Elliot Dec 17 '18 at 9:35
  • $\begingroup$ $I$ stands for integer in my country. $\endgroup$ – kornkaobat Dec 17 '18 at 9:39
  • $\begingroup$ so $a,b,c,d \subseteq \mathbb Z$ if written in international rules. $\endgroup$ – kornkaobat Dec 17 '18 at 9:41
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enter image description here I will not hesitate to put that into mathjax though. For reference only.

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