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

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.

• Is $I$ meant to be the irrational numbers ($\Bbb R - \Bbb Q$) or the integers ($\Bbb Z$)? – Patricio Dec 17 '18 at 8:53
• $$r_1^2=(a-3)^2+(b+4)^2=(a-3)^2+(b-8)^2$$ $\implies b=2,r_1^2=(a-3)^2+(2+4)^2$ 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 $P,a=3$ – lab bhattacharjee Dec 17 '18 at 8:56
• a subset I is meaninglesx. – William Elliot Dec 17 '18 at 9:35
• $I$ stands for integer in my country. – kornkaobat Dec 17 '18 at 9:39
• so $a,b,c,d \subseteq \mathbb Z$ if written in international rules. – kornkaobat Dec 17 '18 at 9:41