Question

Find the equations to the circles in which the line joining the points $(a,b)$ and $(b,-a)$ is a chord subtending an angle of ${45}^{\circ }$ at any point on its circumference.

Answer 1

Let the center of the circle be $(h,k)$
Since the line joining the points $(a,b)$ and $(b,-a)$ subtends ${45}^o$ on the circumference, the chord will subtend ${90}^o$ at the center of the circle.
$\Rightarrow \frac{k-b}{h-a}\times \frac{k+a}{h-b}=-1$
$\Rightarrow k^2-bk+ak-ab=-h^2+ah+bh-ab$
$\Rightarrow k^2+h^2+k(a-b)-h(a+b)=0..\left(1\right)$
Also, $(h-a{)}^2+(k-b{)}^2=(h-b{)}^2+(k+a{)}^2$
$\Rightarrow h^2-2ah+a^2+k^2-2bk+b^2=h^2-2bh+b^2+k^2+2ak+a^2$
$\Rightarrow -2ah-2bk=-2bh+2ak$
$\Rightarrow (b-a)h-(b+a)k=0...\left(2\right)$
From equations (1) and (2), the value of $(h,k)$ can be $(0,0)$ or $(a+b,b-a)$
The equation of circles can therefore be written as $x^2+y^2=a^2+b^2$
or $(x-(a+b){)}^2+(y-(b-a){)}^2=\left[\right(a+b-a{)}^2+(b-a-b{)}^2]$
$\therefore x^2-2(a+b)x+(a+b{)}^2+y^2-2(b-a)y+(b-a{)}^2=a^2+b^2$
i.e. $x^2-2(a+b)x+y^2-2(b-a)y+a^2+b^2=0$