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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.
First problem was Geometry Problem about Cyclic Quadrilateral.

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)

Here is image, if translation is not clear enough.

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

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2 Answers 2

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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 $\frac {b_2}{a_2}$.

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  • $\begingroup$ How to use that fact??? $\endgroup$
    – Tahir Imanov
    Commented Mar 8, 2015 at 9:33
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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

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