Question
$P$ is a point ‘( $a,b$ )’ in the first quadrant. If the two circles which pass through $P$ and touch both the coordinate axes cut at right angles, then:
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Solution
[Key Concept: A circle touching both axes has centre $(r,r)$; through $(a,b)$ gives
$r^{2} - 2(a + b)r + (a^{2} + b^{2}) = 0$, then impose orthogonality of the two such circles.]
A circle touching both axes has centre $(r,r)$: $x^{2} + y^{2} - 2rx - 2ry + r^{2} = 0$; through $(a,b)$ gives $r^{2} - 2(a + b)r + (a^{2} + b^{2}) = 0$, so $r_{1} + r_{2} = 2(a + b),\ r_{1}r_{2} = a^{2} + b^{2}$.
Orthogonality of the two circles: $2(g_{1}g_{2} + f_{1}f_{2}) = c_{1} + c_{2} \Rightarrow \ 4r_{1}r_{2} = r_{1}^{2} + r_{2}^{2} \Rightarrow \ 6r_{1}r_{2} = (r_{1} + r_{2})^{2}$.
Hence $6(a^{2} + b^{2}) = 4(a + b)^{2} \Rightarrow \ a^{2} - 4ab + b^{2} = 0$ → (c).
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