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The radical centre of three circles described on the three sides of a triangle as diameter is
The orthocentre
The circumcentre
The incentre of the triangle
The centroid
Solution
(a) Let us consider a triangle as shown in fig.
Equations of the circles with $AB, BC$ and $CA$ as diameters are ${S_1} \equiv (x + a)(x – a) + {y^2} = 0$
${S_2} \equiv (x – a)(x – \alpha ) + y(y – \beta ) = 0$
and ${S_3} \equiv (x + a)(x – \alpha ) + y(y – \beta ) = 0$
i.e., ${S_1} \equiv {x^2} + {y^2} – {a^2} = 0$
${S_2} \equiv {x^2} + {y^2} – (a + \alpha )x – \beta y + a\alpha = 0$
and ${S_3} \equiv {x^2} + {y^2} – (\alpha – a)x – \beta y – a\alpha = 0$
Radical axis of ${S_2}$ and ${S_3}$ is ${S_3} – {S_2} = 0$
i.e., $2ax – 2a\alpha = 0$
==> $2a(x – \alpha ) = 0$, as $a \ne 0$, $x = \alpha $
But $x = \alpha $ is the orthogonal through $C$.
Similarly other radical axes will be orthogonals through $A$ and $B$.
Hence radical centre will be the orthocentre.
