If the chord $x+y=b$ of the curve ... If the chord $x+y=b$ of the curve $x^2+y^2-2ax-4a^2=0$ subtends a right angle at the origin, prove that: $b(b-a)=4a^2$
My Approach.
Given,
Equation of the chord,
$$x+y=b$$
$$\frac {x+y}{b}=1$$
Now, 
Equation of the curve,
$$x^2+y^2-2ax-4a^2=0$$
$$x^2+y^2-2ax=4a^2$$
$$(b-y)^2+(b-x)^2-2ax=4a^2$$
I got stuck at here. Please help me to complete it.

First solve the equations simultaneously, and we arrive at the quadratic equation $$2x^2-2x(a+b)+b^2-4a^2=0$$
The roots satisfy $$x_1+x_2=a+b$$ and $$x_1x_2=\frac{b^2-4a^2}{2}$$
The perpendicularity condition can be written as $$x_1x_2+y_1y_2=0$$
$$\implies x_1x_2+(b-x_1)(b-x_2)=0$$
$$\implies 2x_1x_2-b(x_1+x_2)+b^2=0$$
Using the above results for the sum and product of the roots, the result follows immediately.
