# Tangent circles and finding diameters

Here is a shaded crescent formed by two internally tangent circles. Point O is the the center of the larger circle. The width at points B and B' is five units. At point A, the width is 9 units. Find the lengths of the two circle's diameters.

I have included the image that was given. Let $r_1$ and $d_1$ refer to the radius and diameter of the larger circle and let $r_2$ and $d_2$ refer to the radius and diameter of the smaller circle.

mBO=$r_1$= x+5

mCO=$r_1$-9

mOD= 4.5, because that is where the center of the smaller circle is. Since the two circles are touching and width of the crescent at A is 9. So, the line drawn in from D would be the radius of the smaller circle.

A right triangle is formed, so I could the use Pythagorean Theorem to find $r_2$.

So:

$r_1$= x+5

$d_1$= 2 x +10

$r_2$= x+4.5

$d_2$= 2 x +9

The problem is that I don't know where to go from here. Am I even on the right track? Any help on how to finish the problem will be appreciated.

I think you have it.

$r_2 = OB = x + 5 = \sqrt{r_1^2 - (9/2)^2} + 5$

$r_2 = OA = 9 + OC = 9 + r_1 - (9/2) = r_1 + 9/2$

So $\sqrt{r_1^2 -(9/2)^2} = r_1 - 1/2 \implies r_1^2 - 81/4 = r_1^2 + 1/4 -r_1 \implies r_1 = 41/2$.

$r_2 = 41/2 + 9/2 = 25$.

$r_2 = r_1 + 4.5$

From the internal right triangle you have:

$(r_2 - 5)^2 + 4.5^2 = r_1^2$.

Solve!

• Disregard my previous comment. Why does $r_2$=$r_1$+4.5? Wouldn't it be $r_2$=$r_1$-4.5? Also, after I find $r_2$, what do I do? – Dana Jan 7 '16 at 3:17
• Dana is right how can you assert$r_2=r_1+4.5$ ? – Mayank Deora Jan 7 '16 at 6:21

Let OC=$a$ and you have $x$ So now we have$$x+5=9+a\tag1$$ and because $$4.5+a=r_2$$By applying pythagorus theorem:- $$(4.5+a)^2-4.5^2=x^2\tag2$$ Now solve equations (1.) and (2.) for the factors $x$ and $a$ and put in the equation :- $$r_1=x+5=9+a$$ and $$r_2=4.5+a$$ There you go........

The easiest way to solve this problem is to use the Intersecting Chord Theorem (ICT). This is one of those simple results, like Pythagoras's Theorem that we learn at school, but then forget.

Looking at the outer circle in the picture below, the ICT tells us, unsurprisingly, that $\|OA\| \times \|OC\| = \|OB\| \times \|OD\|$. Looking at the inner circle in the picture below, the ICT tells us that $\|OE\|\times \|OC\| = \|OF\| \times \|OG\|$.

Assuming that the radius of the outer circle is $r$, we apply the ICT to the inner circle to get: $$\|OE\|\times \|OC\| = \|OF\| \times \|OG\|$$ $$(r-9)\times r = (r-5) \times (r-5)$$ $$r^2-9r = r^2 - 10r + 25$$ $$r=25$$ The radius of the outer circle is $25$, and so it circumference is $50$.

The circumference of the inner circle is $\|EC\|=(r-9)+r=2r-9=41$.