# Geometry Problem about Circles and Tangents

It is the second problem from my maths notebooks, which is still unsolved. I translated it from Russian, so their may be some discrepancies in translation. So, I also added image.

The circle $$S_1$$ with centre $$C_1 (a_1 , b_1)$$ and radius $$r_1$$ touches (or tangent to) externally the circle $$S_2$$ with centre $$C_2 (a_2 , b_2)$$ and radius $$r_2$$. Their common tangent passes through the origin.

Show that, $$(a_1)^2 - (a_2)^2 + (b_1)^2 - (b_2)^2 = (r_1)^2 - (r_2)^2 \; \; \;$$ (i)

If, also other tangents from origin are perpendicular, prove that $$|a_2b_1 -a_1b_2| = |a_1a_2 + b_1b_2| \; \; \;$$ (ii)

I have solved (i), so I need solution only for (ii).

Use the fact that angle $C_1OC_2 = \frac {\pi} 4$

So the angle between x-axis and $OC_1$ is $\frac \pi 4$ more or less than angle between x-axis and $OC_2$

Hint: Use compound angle formulae and $\frac {b_1}{a_1}$ and $\frac {b_2}{a_2}$.

• How to use that fact??? Commented Mar 8, 2015 at 9:33

Continuing from the hint given by $tomi, we have:-$\frac {b_1}{a_1}$is the slope of$OC_1$; and similarly,$\frac {b_2}{a_2}$is the slope of$OC_2\$.

Finding the angle between 2 lines (passing through the origin) with slopes known is given by a standard formula. It can be found in like:-

http://www.vitutor.com/geometry/line/angle_lines.html