Question

If $OA$ and $OB$ are two perpendicular chords of the circle $r=a\cos \theta +b\sin \theta$ passing through origin, then the locus of the mid point of $AB$ is:

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

Answer 1

The correct option is A $x^2+y^2=\frac{ax}{2}+\frac{by}{2}$

$r=a\cos \theta +b\sin \theta$

Multiply by $r$on both sides.

$\Rightarrow r^2=ar\cos \theta +br\sin \theta$

$\because r^2=x^2+y^2$

$r\cos \theta =x$

$r\sin \theta =y$

$\Rightarrow x^2+y^2=ax+by$

$\Rightarrow x^2+y^2-ax-by=0$ (equation of the circle)

Let the mid point of chord $AB$ be $(h,k)$

Equation of chord: $T=S_1$

$\Rightarrow hx+ky-\frac{a}{2}(x+h)-\frac{b}{2}(y+k)=h^2+k^2-ah-bk$

$\Rightarrow hx+ky-\frac{ax}{2}-\frac{by}{2}-h^2+k^2-\frac{ah}{2}-\frac{bk}{2}$

Homogenising:

$\Rightarrow x^2+y^2-(ax+by)\left(\frac{(hx+ky-\frac{ax}{2}-\frac{by}{2})}{(h^2+k^2-\frac{ah}{2}-\frac{bk}{2})}\right)=0$

Let $\text{coff of xy}$ be $p$

$\Rightarrow x^2(h^2+k^2-\frac{ah}{2}-\frac{bk}{2})+y^2(h^2+k^2-\frac{ah}{2}-\frac{bk}{2})-(ah{x}^2-\frac{a^2{x}^2}{2}+bk{y}^2-\frac{b^2{y}^2}{2})+xy\left(p\right)=0$

$\Rightarrow m_1{m}_2=\frac{a}{b}=-1$

$\Rightarrow h^2+k^2-\frac{ah}{2}-\frac{bk}{2}=-h^2-k^2+\frac{ah}{2}+\frac{bk}{2}$

$\Rightarrow 2(h^2+k^2-\frac{ah}{2}-\frac{bk}{2}=0$

Locus : $x^2+y^2-\frac{ax}{2}-\frac{by}{2}=0$