The radical centre of three circles described | vedclass

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Question

(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^

Vedclass · Accepted Answer

The orthocentre

Vedclass · Answer

(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 $a 
e 0$, $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.

The radical centre of three circles described on the three sides of a triangle as diameter is

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