Question

Circles are drawn on the three sides of a triangle as diameters. Prove that their radical centre is the orthocentre of the triangle.

Answer 1

Choose the side $BC=2a$ along x-axis and its mid-point be taken as origin so that $B$ and $C$ are as marked. Let $A$ be $(h,k)$. Circles on sides as diameters are
$S_1=x^2+y^2-a^2=0$
$S_2=(x-h)(x-a)+(y-k)y=0$
or $x^2+y^2-x(h+a)-ky+ah=0$
Replacing $a$ by $-a$
$S_3=x^2+y^2-x(h-a)-ky-ah=0$
Radical axis of $S_1$ and $S_2$ is $S_1-S_2=0$
$x(h+a)+ky-ah-a^2=0$
Replacing $a$ by $-a$ radical axis of $S_1$ and $S_3$ is
$x(h-a)+ky+ah-a^2=0$
Solving the above two we get the radical centre as $(h,\frac{a^2-h^2}{k})$
Now in order to find orthocentre we have to find the equation of altitudes $AD$ and $BE$.
$AD$ is $x=h$
$\therefore BE:y-0=-\frac{h+a}{k}(x-a)\because BE\perp AC$
Solving we get the orthocentre as $(h,\frac{a^2-h^2}{k})$
which is same as radical centre.