<s>[INST] Circle and Locus _ ONLY PEN AND PAPER ALLOWED. Q) Let T be the line passing through the points P(–2, 7) and Q(2, –5). Let $F_{1}$ be the set of all pairs of circles $(S_{1}$, $S_{2}$) such that T is tangent to $S_{1}$ at P and tangent to $S_{2}$ at Q, and also such that $S_{1}$ and $S_{2}$ touch each other at a unique point, say, M. Let $E_{1}$ be the set representing the locus of M as the pair ($S_{1}$, $S_{2}$) varies in $F_{1}$. Let the set of all straight line segments joining a pair of distinct points of $E_{1}$ and passing through the point R(1, 1) be $F_{2}$. Let $E_{2}$ be the set of the mid-points of the line segments in the set $F_{2}$. Let $C$ be the circle $x^2+y^2+6(2y+7x)=53$. The number of times $C$ intersects $E_{1}$ and $E_{2}$ is (are): 
 [/INST] As the diameter of the first is $PQ$ $$E_1=\{(x,y)| (x+2)(x-2)+(y-7)(y+5)=0\}-\{P,Q\}$$
and the diameter of the second is $DR$, with $D$ the midpoint of $PQ$
$$E_2=\{(x,y)| x(x-1)+(y-1)^2=0\}-\{(4/5,7/5),(36/37,43/37)\}$$
where the disallowed points stem from the intersections with $l_{PR}$ and $l_{QR}$.
Now
$$\#((V((x+2)(x-2)+(y-7)(y+5))\cup V(x(x-1)+(y-1)^2))\cap C)=4$$
but three points are disallowed, leaving the other intersection $(400/409,349/409)$ so
$$\#((E_1\cup E_2)\cap C)=1$$

Edit
$E_2$ is part of a circle: let $y-1=m(x-1)$ be lines through $R$, then together with the equation for the circle $E_1$ is a part of, we get the relation $(m^2+1)y^2+(-2m^2+2m-2)y-38m^2-2m+1=0$ that gives us solutions $y_{1,2}$ with corresponding $x_{1,2}$. What we're looking for are points $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{1}{m^2+1},\frac{m^2-m+1}{m^2+1})$ which implicitizes to $x^2-x+y^2-2y+1=0.$
</s>