Question

Two line segments $AB$ and $CD$ are constrained to move along the $x$ and $y$ axes, respectively, in such a way that points $A,B,C,D$ are concyclic. If $AB=a$ and $CD=b$, then the locus of the centre of the circle passing through $A,B,C,D$ in polar coordinates is:

• A. $r^2=\frac{a^2+b^2}{4}$
• B. $r^2\cos 2\theta =\frac{a^2-b^2}{4}$
• C. $r^2=4(a^2+b^2)$
• D. $r^2\cos \theta =4(a^2-b^2)$

Answer 1

The correct option is B $r^2\cos 2\theta =\frac{a^2-b^2}{4}$

Observe the figure above.

Let the coordinates of center of the circle $O$ be $(x,y)$

Drop the perpendiculars from $O$ onto the segments $AB$ and $CD$.

From the figure, $x$ = length of the perpendicular to $CD$

i.e., $|x|=\sqrt{d^2-(b/2{)}^2}$ ---------(1)

Similarly, $y$ = length of the perpendicular to $AB$

i.e., $|y|=\sqrt{d^2-(a/2{)}^2}$ ----------(2)

Eliminating $d$ from (1) and (2),

$x^2-y^2=(d^2-(b/2{)}^2)-(d^2-\left(a/2{)}^2\right)=\frac{a^2-b^2}{4}$

But, in polar coordinates, $x=r\cos \theta$ and $y=r\sin \theta$

Thus, $x^2-y^2=r^2({\cos }^2\theta -{\sin }^2\theta )=r^2\cos 2\theta$

Thus, the polar equation for the locus of $O$ is $r^2\cos 2\theta =\frac{a^2-b^2}{4}$