Let $C_1$ be a circle defined with $X(-2,7)$ and $Y(2,-5)$ as the endpoints of the diameter of the circle.
Let $C_2$ be a circle defined with $Y(2,-5)$ and $Z(4,-11)$ as the endpoints of the diameter of the circle.
Let $A = C_1 \cup C_2 – Y$
Let $B$ be the set of all midpoints of the set of all straight line segments joining two distinct points in $A$ and passing through the point $Y$. Find $B$.
My attempt:
Since $XYZ$ are collinear $C_1$ and $C_2$ meet at a single point: $Y$. I can easily find the set of all midpoints of the line segments joining a point (say $T$) in $C_1$ and $Y$. Let $M(x,y)$ be the midpoint of $TY$ and $N(0,1)$ be the center of $C_1$. Notice that $MN$ is perpendicular to to $MY$. Therefore we can build the equation $\dfrac{y-1}{x-0}*\dfrac{y+5}{x-2}=-1$ i.e product of their slopes equal -1. Similarly, the set of all midpoints of the line segments joining a point (say $U$) in $C_2$ and $Y$ can be found.
The issue now is, I don't know how to use this information to compute the required locus $B$. I don't even know if this information is useful.
Best Answer
Notice that transformation $A\mapsto B$ from bigger circle which has radius $2\sqrt{10}$: $$XY = \sqrt{4^2 +12^2} = 4\sqrt{10}$$ and center at $O(0,1)$ to smaller circle with radius $\sqrt{10}$ and center at $O'(3,-8)$ is homothety with center at $Y$ and dilatation factor $-1/2$. So, for all $A$ we have $${BY\over YA} = {1\over 2}\implies {AB\over YA} ={3\over 2}$$
Then $${YM \over YA} = 1- {AM\over YA} = 1-{AB\over 2YA} = {1\over 4} $$
So $M$ is actually a picture of $A$ under homothety at $Y$ with dilatation factor $1/4$, so $M$ descibes a circle with radius $r=\sqrt{10}/2$ and center at $({1\over 2},-{7\over 2})$.