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.

The combined equation of the lines joining the origin to the end points of the chord can be obtained by "homogenising" the equation of the curve. This is 
\begin{align*}
x^2+y^2 - 2ax\left(\frac{x+y}{b}\right) - 4a^2\left(\frac{x+y}{b}\right)^2 = 0 
\end{align*}
These lines are perpendicular if the sum of the coefficients of $x^2$ and $y^2$ is zero. Thus the required condition is 
\begin{align*}
1+1-\frac{2a}{b}-\frac{4a^2}{b^2}(1+1) &= 0\\
b^2 - ab -4a^2 &= 0\\
b(b-a) &= 4a^2
\end{align*}
