Locus of Contact Points of Tangent Circles and Midpoint Sets

To determine the correct statements, we analyze the geometric conditions step-by-step.

Step 1: Finding the Locus $E_1$

Let $P = (-2, 7)$ and $Q = (2, -5)$ be points on line $T$. The slope of $T$ is: $m = \frac{-5 - 7}{2 - (-2)} = \frac{-12}{4} = -3$ The equation of line $T$ is: $y - 7 = -3(x + 2) \implies 3x + y - 1 = 0$

Let $S_1$ and $S_2$ be circles tangent to line $T$ at $P$ and $Q$ respectively, touching each other at point $M$.

The common tangent to $S_1$ and $S_2$ at $M$ intersects line $T$ at a point $K$. Since the lengths of tangents from $K$ to $S_1$ and $S_2$ are equal: $KP = KM = KQ$

Thus, $K$ is the midpoint of segment $PQ$: $K = \left( \frac{-2 + 2}{2}, \frac{7 - 5}{2} \right) = (0, 1)$

The distance $KP$ is given by: $KP = \sqrt{(0 - (-2))^2 + (1 - 7)^2} = \sqrt{4 + 36} = \sqrt{40}$

Since $KM = KP = \sqrt{40}$, the point $M$ lies on the circle centered at $K(0, 1)$ with radius $\sqrt{40}$: $x^2 + (y - 1)^2 = 40$

Since $S_1$ and $S_2$ are distinct circles tangent to $T$ at $P$ and $Q$, $M$ cannot coincide with $P$ or $Q$. Thus, $E_1$ is the set of all points on the circle $x^2 + (y - 1)^2 = 40$ excluding the points $P(-2, 7)$ and $Q(2, -5)$: $E_1 = \{ (x, y) \mid x^2 + (y - 1)^2 = 40 \} \setminus \{(-2, 7), (2, -5)\}$

• Option A Analysis: Since $P(-2, 7) \notin E_1$, statement (A) is FALSE.
• Option D Analysis: Testing the point $\left(0, \frac{3}{2}\right)$: $0^2 + \left(\frac{3}{2} - 1\right)^2 = \frac{1}{4} \neq 40$ Since $\left(0, \frac{3}{2}\right)$ does not lie on the circle, it does NOT lie in $E_1$. Hence, statement (D) is TRUE.

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Step 2: Finding the Locus of Midpoints $E_2$

$F_2$ is the set of line segments connecting distinct points $A, B \in E_1$ such that the line segment $AB$ passes through $R(1, 1)$.

The midpoint $M_0(x, y)$ of a chord $AB$ of the circle $x^2 + (y - 1)^2 = 40$ passing through $R(1, 1)$ satisfies $KM_0 \perp AB$. Thus, $\angle K M_0 R = 90^\circ$, which means $M_0$ lies on the circle with diameter $KR$, where $K = (0, 1)$ and $R = (1, 1)$: $(x - 0)(x - 1) + (y - 1)(y - 1) = 0 \implies x^2 - x + (y - 1)^2 = 0$

• Option C Analysis: Testing $\left(\frac{1}{2}, 1\right)$ in the equation of the midpoint circle: $\left(\frac{1}{2}\right)^2 - \frac{1}{2} + (1 - 1)^2 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \neq 0$ Since $\left(\frac{1}{2}, 1\right)$ is the center of the midpoint circle, it does not lie on the boundary where all midpoints lie. Thus, statement (C) is FALSE.

• Option B Analysis: Consider the line passing through $P(-2, 7)$ and $R(1, 1)$. Its slope is $\frac{1 - 7}{1 - (-2)} = -2$, and its equation is: $y - 1 = -2(x - 1) \implies 2x + y - 3 = 0$

  Substituting $y - 1 = 2 - 2x$ into the circle equation $x^2 + (y - 1)^2 = 40$: $x^2 + (2 - 2x)^2 = 40 \implies 5x^2 - 8x - 36 = 0$ Solving for $x$ gives $x = -2$ (which corresponds to point $P$) and $x = \frac{18}{5}$. The second intersection point is $P'\left(\frac{18}{5}, -\frac{21}{5}\right)$.

  setup the midpoint of $PP'$: $x_{mid} = \frac{-2 + \frac{18}{5}}{2} = \frac{4}{5}, \quad y_{mid} = \frac{7 - \frac{21}{5}}{2} = \frac{7}{5}$

  Since $P(-2, 7) \notin E_1$, the segment $PP'$ is excluded from $F_2$. Because a chord through $R(1, 1)$ with midpoint $\left(\frac{4}{5}, \frac{7}{5}\right)$ must uniquely have $P$ as one of its endpoints, no line segment in $F_2$ has $\left(\frac{4}{5}, \frac{7}{5}\right)$ as its midpoint. Thus, $\left(\frac{4}{5}, \frac{7}{5}\right) \notin E_2$, making statement (B) TRUE.

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Conclusion

The correct statements are B and D.