The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta, 2\sin \theta)$ internally in the ratio $2 : 3$ for all $\theta$ is a

The locus of the centre of circles passing through $(a, b)$ and cutting the circle $x^2+y^2-2x+4y-4=0$ orthogonally is

Find the locus of the midpoint of a chord of the circle $x^2 + y^2 = a^2$ which subtends a right angle at the center.

Let $P$ be a point on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line through $P$ parallel to the $Y$-axis meets the circle $x^{2}+y^{2}=9$ at $Q$, where $P$ and $Q$ are on the same side of the $X$-axis. If $R$ is a point on $PQ$ such that $\frac{PR}{RQ}=\frac{1}{2}$, then the locus of $R$ is

Let $RS$ be the diameter of the circle $x^2+y^2=1$,where $S$ is the point $(1,0)$. Let $P$ be a variable point (other than $R$ and $S$) on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q$. The normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $RS$ at point $E$. Then the locus of $E$ passes through the point$(s)$:
$(A)$ $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$ $(B)$ $\left(\frac{1}{4}, \frac{1}{2}\right)$ $(C)$ $\left(\frac{1}{3},-\frac{1}{\sqrt{3}}\right)$ $(D)$ $\left(\frac{1}{4},-\frac{1}{2}\right)$

If a circle passes through the point $(a, b)$ and cuts the circle $x^2 + y^2 = K^2$ orthogonally,then the equation of the locus of its centre is: