# Thread: three circles inscribed in a big circle

1. ## three circles inscribed in a big circle

hello ! help me to solve this problem, please!!
its really hard for me ! thank you so much and God bless you richly !

Three circles A,B and C are tangent externally to each other and each tangent internally to a larger circle having a radius of 10 cm. Radius of circle A is 5 cm.

a) compute the distance from the center of the larger circle to the point of tangency of the two circles B and C which are identical.

b) compute the radius of circles B and C.

2. Hi arenkun,

For the radii, you can use a combination of Sin(angle) and the Cosine Rule
(Law of Cosines) to solve this.

In the diagram...

$\displaystyle \vartriangle{oc_1c_2}$ has sides 5, 5+r, 10-r.

Also $\displaystyle \sin\theta = \frac{r}{5+r}$

From the Law of Cosines...

$\displaystyle (10-r)^2=(5+r)^2+5^2-2(5)(5+r)\cos\theta$

$\displaystyle 100-20r+r^2=25+10r+r^2+25-10(5+r)\cos\theta$

$\displaystyle 50-30r=-(50+10r)\cos\theta$

$\displaystyle \cos\theta = \frac{50-30r}{-(50+10r)}=\frac{30r-50}{50+10r}$

Now use the trigonometric identity $\displaystyle Sin^{2}\theta + Cos^{2}\theta = 1.$

$\displaystyle \frac{(30r-50)^2}{(50+10r)^2}+\frac{r^2}{(5+r)^2} = 1$

$\displaystyle \frac{900r^2-3000r+2500}{2500+100r^2+1000r} + \frac{r^2}{25+10r+r^2}=1$

$\displaystyle \frac{9r^2-30r+25}{25+r^2+10r}+\frac{r^2}{25+10r+r^2}=1$

$\displaystyle 10r^2-30r+25=r^2+10r+25$

$\displaystyle 9r^2-40r=0$

$\displaystyle r(9r-40)=0$

As r is not zero, then $\displaystyle 9r=40$

$\displaystyle r=\frac{40}{9}$

3. Once you have "r", you can use it to calculate the answer
to the first question as shown on the attachment.

4. However, you may have meant the large circle to be the green one!

Let us know if you did.