If the sum of the distances of a point from the origin and the line $x = 2$ is $4$,then its locus is

$A$ circle is drawn to cut a chord of length $2a$ units along the $X$-axis and to touch the $Y$-axis. The locus of the centre of the circle 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 $t \in R - \{-1\}$,then the locus of the point $\left(\frac{3at}{1+t^3}, \frac{3at^2}{1+t^3}\right)$ is

Find the locus of a point such that the sum of the squares of its distances from the axes is $4$.

The locus of the centre of the circle,which cuts the circle $x^2+y^2-20x+4=0$ orthogonally and touches the line $x=2$,is