<s>[INST] 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.
 [/INST] 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*}
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