# three circles inscribed in a big circle

• Jan 12th 2010, 05:24 AM
arenkun
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 !(Wink)

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.
• Jan 12th 2010, 08:10 AM
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...

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

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

From the Law of Cosines...

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

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

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

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

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

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

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

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

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

$9r^2-40r=0$

$r(9r-40)=0$

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

$r=\frac{40}{9}$
• Jan 12th 2010, 12:19 PM
Once you have "r", you can use it to calculate the answer
to the first question as shown on the attachment.
• Jan 12th 2010, 12:23 PM
However, you may have meant the large circle to be the green one!

Let us know if you did.
• Jan 12th 2010, 06:51 PM
arenkun