Intersecting circles: finding the centre of one using the other? Hi guys I'm working out a problem which needs me to solve these circle equations but it's been a while since I've had to do this stuff. I need to find the y coordinate of the red circle radius r/4 when its x-coord is r/4 and is also meeting the blue circle radius r.
It seems to me that its possible to work this out with only the given information but am I wrong?

I've tried using both cartesian and parametric equations but it comes out incorrectly.
e.g.
$$
(x-\frac{r}{4})^2 + (y-b)^2 - \frac{r^2}{16} = x^2 + y^2 - r^2
$$
$$
x  -\frac{r}{4} + y -b = x + y - \sqrt{\frac{15}{16}}r
$$
$$
-\frac{r}{4} = - \sqrt{\frac{15}{16}}r +b
$$
$$
(\sqrt{\frac{15}{16}}-\sqrt{\frac{1}{16}})r =  b
$$
$$
\frac{\sqrt{15}-1}{4} = b
$$
I'm guessing I might not be allowed to remove x and y or I'm simplifying incorrectly..
Thanks, Dan
 A: Nice diagram, it made everything clear. Unfortunately, I will not reciprocate, and will use words to identify certain key points in the picture. 
Let $C$ be the centre of the little circle. Let the origin, and centre of the big circle, be $O$.  
Draw the line $OC$. Extended, it goes through the point $P$ of tangency of the two circles. This is because the common tangent line through $P$ is simultaneously perpendicular to the radii $OP$ and $CP$.  
Draw the vertical line through $O$. This is tangent to the little circle. Draw the line through $C$ perpendicular to this vertical line. It meets the vertical line at the point $T$ of tangency with the little circle.
Consider $\triangle OTC$. We have $TC=\frac{r}{4}$. Also, $OC=\frac{3}{4}r$. This is because $OC=OP-CP=r-\frac{r}{4}$.
Thus by the Pythagorean Theorem
$$OT^2=\frac{9}{16}r^2-\frac{1}{16}r^2,$$
and therefore $OT=\frac{1}{\sqrt{2}}r$. This is the desired $y$-coordinate of the centre of the little circle.  
Remark: We can also set up and solve suitable equations, For this problem, that approach is somewhat more complicated, certainly more complicated to type. 
