# Circles

• Oct 7th 2009, 09:22 PM
arze
Circles
The circle $S_1$ with center $C_1 (a_1,b_1)$ and radius $r_1$ touches externally the circle $s_2$ center $C_2 (a_2,b_2)$ radius $r_2$. The tangent at their common point passes through the origin. Show that,
$({a_1}^2-{a_2}^2)+({b_1}^2-{b_2}^2)=({r_1}^2-{r_2}^2)$
If also, the other two tangents from the origin to $S_1$ and $S_2$ are perpendicular, prove that $|a_2b_1-a_1b_2|=|a_1a_2+b_1b_2|$.
Hence show that if $C_1$ remains fixed but $S_1$ and $S_2$ vary, then $C_2$ lies on the curve
$({a_1}^2-{b_1}^2)(x^2-y^2)+4a_1b_1xy=0$.
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.
• Oct 7th 2009, 10:50 PM
pacman
Arze: "The circle http://www.mathhelpforum.com/math-he...5727065b-1.gif with center http://www.mathhelpforum.com/math-he...56adc012-1.gif and radius http://www.mathhelpforum.com/math-he...34c06429-1.gif touches externally the circle http://www.mathhelpforum.com/math-he...0cc2fbae-1.gif center http://www.mathhelpforum.com/math-he...4944782c-1.gif radius http://www.mathhelpforum.com/math-he...8a8e9448-1.gif. The tangent at their common point passes through the origin. Show that,
http://www.mathhelpforum.com/math-he...2bfc64ed-1.gif"

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)
• Oct 8th 2009, 05:47 PM
arze
but how do we know that the point is (0,0)?