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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?

enter image description here

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

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up vote 4 down vote accepted

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.

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Thanks a lot, a much simpler method and it works :). Was that simultaneous equation a possible way of doing it? –  dnv Dec 27 '12 at 15:32
    
Equations will work. One needs a "circles touch at $1$ point only" equation. This involves a discriminant. Your calculation had an immediate error in going from first displayed line to the second. You assumed that $\sqrt{s^2+t^2}=s+t$. This is not true. –  André Nicolas Dec 27 '12 at 15:43
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