1. ## circles in circles

The center and radius of circumference M are (0,0) and 10, respectively. The center and radius of circumference P are (8,0) and 2, respectively. Circumference Q is internally tangent to circumference M and externally tangent to circumference P.Of course that there are infinitely many circumferences Q which have this property. Determine and graph the equation of the curve that contains the centers of all the circumferences of the Q kind.

$\displaystyle x^{2}\;+\;y^{2}\;=\;10^{2}$

$\displaystyle (x-8)^{2}\;+\;y^{2}\;=\;2^{2}$

$\displaystyle (x+2)^{2}\;+\;y^{2}\;=\;8^{2}$

What's your plan for the rest?

3. Originally Posted by rlarach
The center and radius of circumference M are (0,0) and 10, respectively. The center and radius of circumference P are (8,0) and 2, respectively. Circumference Q is internally tangent to circumference M and externally tangent to circumference P.Of course that there are infinitely many circumferences Q which have this property. Determine and graph the equation of the curve that contains the centers of all the circumferences of the Q kind.

Hello,

I've attached a drawing of the problem:

1. the blue circles are your circles P and M
2. the red circles are a few examples of circle Q
3. the black curve is the path of all centres of Q

By pure guessing I found out that this curve must be an ellipse with the equation:

$\displaystyle \frac{(x-4)^2}{6^2} + \frac{y^2}{\left(\frac92 \right)^2} = 1$

4. Originally Posted by rlarach
The center and radius of circumference M are (0,0) and 10, respectively. The center and radius of circumference P are (8,0) and 2, respectively. Circumference Q is internally tangent to circumference M and externally tangent to circumference P.Of course that there are infinitely many circumferences Q which have this property. Determine and graph the equation of the curve that contains the centers of all the circumferences of the Q kind.

Tätäää!

I finally found a solution.

I've attached a drawing of the problem.

Let C(a, b) be the centre of the circle Q. The you have 2 right triangles (coloured grey and light blue). Use Pythagorean theorem:

1. Grey triangle: $\displaystyle a^2+b^2=(10-r)^2$
2. Blue triangle: $\displaystyle (8-a)^2+b^2 = (r+2)^2$

I solved this system of equations for the variable r and plugged the result into the equation 1.:

$\displaystyle \boxed{\begin{array}{lcr}a^2+b^2 &=& 100-20r+r^2 \\64-16a+a^2+b^2 &=& 4+4r+r^2\end{array}}$. Now subtract the second equation from the first on:

$\displaystyle -64+16a=96-24r$ . Solve for r:

$\displaystyle r=\frac{20}{3}-\frac23 a$ . I plugged this value into the first equation:

1. Grey triangle: $\displaystyle a^2+b^2=\left(10-\left(\frac{20}{3}-\frac23 a\right) \right)^2$. Expand the bracket, rearrange the equation and multiply by $\displaystyle \frac95$. You'll get:

$\displaystyle a^2-8a+\frac95 b^2=20$ . Complete the square:

$\displaystyle a^2-8a+16+ \frac95 b^2=20+16$ and divide by 36:

$\displaystyle \frac{(a-4)^2}{6^2}+\frac{y^2}{20}=1$

This is the the equation of an ellipse with the centre at (4, 0) and the axes: 6 and $\displaystyle \sqrt{20} \approx 4.472$. Therefore my previous post is not correct: I said that the minor axis has a length of 4.5. Sorry.

5. Please don't double post. See rule #1 here.

This is why we don't double post: ticbol answered your post in the other thread and earboth did it here. This is a waste of the time of the members that are trying to help. Please don't do it again!

-Dan