Question

Prove that the equations $r=a\cos (\theta -\alpha )$ and $r=b\sin (\theta -\alpha )$ represent two circles which cut at right angles.

Answer 1

Condition for a two circle to cut at right angle $\Rightarrow (Distancebetweencentres{)}^2=radiu{s}_1^2+radiu{s}_2^2$

First Circle-

$r=acos(\theta -\alpha )$

Substitute

$x=rcos\theta$

$y=rsin\theta$

$r=acos\theta cos\alpha +asin\theta sin\alpha$

$r=\frac{a}{r}xcos\alpha +\frac{a}{r}ysin\alpha$

$r^2=axcos\alpha +aysin\alpha$

Replacing $rby{x}^2+y^2$

$x^2+y^2=axcos\alpha +aysin\alpha$

${(x-\frac{a}{2}cos\alpha )}^2+{(y-\frac{a}{2}sin\alpha )}^2={\left(\frac{a}{2}\right)}^2$

Second circle

$r=asin(\theta -\alpha )$

$r=asin\theta cos\alpha -asin\alpha cos\theta$

put $x=rcos\theta$

$y=rsin\theta$

$\Rightarrow r=\frac{a}{r}ycos\alpha -\frac{a}{r}xsin\alpha$

$r^2=aycos\alpha -axsin\alpha$

$x^2+y^2=aycos\alpha -axsin\alpha$

${(x+\frac{a}{2}sin\alpha )}^2+{(y-\frac{a}{2}cos\alpha )}^2={\left(\frac{a}{2}\right)}^2$

Square of distance between 2 centres $={(\frac{a}{2}cos\alpha +\frac{a}{2}sin\alpha )}^2+{(\frac{a}{2}sin\alpha -\frac{a}{2}cos\alpha )}^2$

On solving we get it equal to $\frac{a^2}{2}$

Sum of squares of radius = $\frac{a^2}{4}+\frac{a^2}{4}$ which also equals $\frac{a^2}{2}$

Hence proved, the circles intersect at right angles.