Question

The locus of the midpoints of the chords of the circle $x^2+y^2-ax-by=0$ which subtend a right angle at $(\frac{a}{2},\frac{b}{2})$ is

• A. $ax+by=0$
• B. $ax+by=a^2+b^2$
• C. $x^2+y^2-ax-by+\frac{a^2+b^2}{8}=0$
• D. $x^2+y^2-ax-by-\frac{a^2+b^2}{8}=0$

Answer 1

Answer: (D)

$x^2+y^2-ax-by-\frac{a^2+b^2}{8}=0$

Centre of the circle $=(\frac{a}{2},\frac{b}{2})$

Radius of the circle $=\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}=\frac{\sqrt{a^2+b^2}}{2}$

$\sin {45}^o=\frac{\sqrt{{(h-\frac{a}{2})}^2+{(k-\frac{b}{2})}^2}}{\frac{\sqrt{a^2+b^2}}{2}}$

$\Rightarrow \frac{1}{2}=4\left|\frac{\frac{(2h-a{)}^2}{4}+\frac{(2k-b{)}^2}{4}}{a^2+b^2}\right|$

Simplify to get locus $x^2+y^2-ax-by-\frac{a^2+b^2}{8}=0$