Question

The locus of the mid points 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. $ax+by=a^2+b2$
• 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

The correct option is C

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

$C=(\frac{a}{2},\frac{1}{2}),r=\frac{1}{2}\sqrt{a^2+b^2}$

P = (x, y)

$cos\frac{\pi }{4}=\frac{CP}{r}\Rightarrow C{P}^2=r^{2}co{s}^2\frac{\pi }{4}{(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2=\frac{a^2+b^2}{8}$