# Thread: Circles

1. ## Circles

The circle $\displaystyle S_1$ with center $\displaystyle C_1 (a_1,b_1)$ and radius $\displaystyle r_1$ touches externally the circle $\displaystyle s_2$ center $\displaystyle C_2 (a_2,b_2)$ radius $\displaystyle r_2$. The tangent at their common point passes through the origin. Show that,
$\displaystyle ({a_1}^2-{a_2}^2)+({b_1}^2-{b_2}^2)=({r_1}^2-{r_2}^2)$
If also, the other two tangents from the origin to $\displaystyle S_1$ and $\displaystyle S_2$ are perpendicular, prove that $\displaystyle |a_2b_1-a_1b_2|=|a_1a_2+b_1b_2|$.
Hence show that if $\displaystyle C_1$ remains fixed but $\displaystyle S_1$ and $\displaystyle S_2$ vary, then $\displaystyle C_2$ lies on the curve
$\displaystyle ({a_1}^2-{b_1}^2)(x^2-y^2)+4a_1b_1xy=0$.
I'm quite sure if I know how to do the first part, I should be able to do the rest, but I don't even know how to start. Thanks for any help.

2. Arze: "The circle with center and radius touches externally the circle center radius . The tangent at their common point passes through the origin. Show that,
"

Solution for (i) if they have a common point of (0,0), use the standard formula for circle:

(0 - A)^2 + (0 - B)^2 = R^2,
(0 - a)^2 + (0 - b)^2 = r^2,

subtract,

(A^2 - a^2) + (B^2 - b^2) = (R^2 - r^2)

3. but how do we know that the point is (0,0)?