If the coordinates of a point are given by the equations $x = a(1 - \cos \theta )$ and $y = a\sin \theta $,then the locus of the point will be

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$(s)$:
$(A)$ $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$ $(B)$ $\left(\frac{1}{4}, \frac{1}{2}\right)$ $(C)$ $\left(\frac{1}{3},-\frac{1}{\sqrt{3}}\right)$ $(D)$ $\left(\frac{1}{4},-\frac{1}{2}\right)$

The locus of the mid-points of the chords of the circle $x^{2}+y^{2}+2x-2y-2=0$ which make an angle of $90^{\circ}$ at the centre is

The locus of the centers of the circles touching the lines $3x - 4y + 1 = 0$ and $12x + 5y - 1 = 0$ is/are:

$A$ point $P$ moves such that the distance from $(0,2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1,0)$. Then the locus of the point is

$A$ stick of length $r$ units slides with its ends on coordinate axes. Then the locus of the midpoint of the stick is a curve whose length is