Question

Line $y=x$ cuts the curve $y^2=4ax$ at $A$ and $B$. Find equation of circle with $AB$ as diameter.

Answer 1

$y=x$ & $y^2=4ax$

$x^2=4ax$

$x=0$ or $x=4a$

$x=0,y=0$ and $x=4a,y=4a$

$A\equiv (0,0)$ $B\equiv (4a,4a)$

equation of circle with A & B as end points of diameter $\to$

$(x-0)(x-4a)+(y-0)(y-4a)=0$

$x^2-4ax+y^2-4ay=0$