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.
On
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.
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*}