Question

Find the radius of a circle $(xcos\theta +ysin\theta -a{)}^2+(xsin\theta +ycos\theta -b{)}^2=k^2$
If $\theta$ varies then prove that locus of the centre of above circle is again a circle.

Answer 1

The given circle is
x${}^2$2.1 + y${}^2$2.1 - 2x(a cos $\theta$ + b sin $\theta$)
-2y(a sin $\theta$ - b cos $\theta$) + (a${}^2$ + b${}^2$ k${}^2$) = 0
Above equation represents a circle whose centre is the point
x = a cos $\theta$ + b sin $\theta$
y = a sin $\theta$ - b cos $\theta$ where $\theta$ varies.
The locus of centre is obtained by eliminating the variable $\theta$. Squaring and adding, the required locus is x${}^2$ + y${}^2$ = a${}^2$ + b${}^2$
Also radius of given circle is
$\sqrt{g^2+f^2-c}=\sqrt{k^2}=k$