Question

Points $(1,0)$ and $(2,0)$ are taken on the axis of $x$, the axes being rectangular. On the line joining these points an equilateral triangle is described, its vertex being in the positive quadrant. Find the equations to the circles described on its sides as diameters.

Answer 1

this is an equilateral triangle so,

$a=b=c$

$(h-1{)}^2+k^2=(h-2{)}^2+k^2=1$

$h=\frac{3}{2},k=\frac{\sqrt{3}}{2}$

Equation of circle with sides as diameter is given by

On side $AB$

$(x-1)(x-2)+(y-0)(y-0)=0$

$x^2+y^2-3x+2=0$

On side $BC$

$(x-2)(x-\frac{3}{2})+(y-0)(y-\frac{\sqrt{3}}{2})=0$

$2x^2+2y^2-7x-\sqrt{3}y+6=0$

On side $AC$

$(x-1)(x-\frac{3}{2})+(y-0)(y-\frac{\sqrt{3}}{2})=0$

$2x^2+2y^2-5x-\sqrt{3}y+3=0$