For any real number $t$,the point $\left(\frac{8 t}{1+t^2}, \frac{4\left(1-t^2\right)}{1+t^2}\right)$ lies on a/an

Find the locus of a point from which the lengths of the tangents drawn to the two circles $x^2 + y^2 - 5x - 3 = 0$ and $3x^2 + 3y^2 + 2x + 4y - 6 = 0$ are equal.

Let $PQ$ and $MN$ be two straight lines touching the circle $x^{2}+y^{2}-4x-6y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle AOB=\pi/3$. Then the locus of the point of intersection of the lines $PQ$ and $MN$ 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$(s)$:
$(A)$ $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$ $(B)$ $\left(\frac{1}{4}, \frac{1}{2}\right)$ $(C)$ $\left(\frac{1}{3},-\frac{1}{\sqrt{3}}\right)$ $(D)$ $\left(\frac{1}{4},-\frac{1}{2}\right)$

The locus of the midpoints of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the origin is

$A$ company situated at $(2,0)$ in the $XY$-plane charges $RS. 2$ per $km$ for delivery. $A$ second company at $(0,3)$ charges $RS. 3$ per $km$ for delivery. The region of the plane where it is cheaper to use the first company is