Midpoint of a line segment passing through two points.

Question

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 $B$.

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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.

Answer 1

Locus of points dividing segments joining points of $S_1$ with points of $S_2$ in ratio $\lambda : (1-\lambda)$ is given by $$\lambda S_1 + (1-\lambda) S_2=0$$

$(2,-5)$ should lie on this locus, which it clearly does. For midpoints, put $\lambda = \frac{1}{2}$. The coefficient of $x^2$ and $y^2$ will turn out to be same; this locus is a circle.

Given $$S_1 : (x-2)(x+2) + (y+5)(y-7)=0$$ $$S_2 : (x-2)(x-4) + (y+5)(y+11)=0$$

Required locus is $$S : (x-2)(x-1) + (y+5)(y+2)=0$$

with center at $$\Big(\dfrac{2+1}{2},\dfrac{-5-2}{2}\Big)=\Big(\dfrac{3}{2},\dfrac{-7}{2}\Big)$$

Answer 2

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 $({1\over 2},-{7\over 2})$.