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(D) Let $M \equiv (h, k)$. Since $MP$ and $MQ$ are tangents from $M$ to the circles $S_1$ and $S_2$ respectively,and $M$ is the point of contact,$MP =

Vedclass · Accepted Answer

The point $(0, \frac{3}{2})$ does $NOT$ lie in $E_1$

Vedclass · Answer

(D) Let $M \equiv (h, k)$. Since $MP$ and $MQ$ are tangents from $M$ to the circles $S_1$ and $S_2$ respectively,and $M$ is the point of contact,$MP = MQ$. Also,the angle $\angle PMQ = 90^{\circ}$.Thus,the slope of $MP 	imes$ slope of $MQ = -1$.$\left(\frac{k-7}{h+2}ight) 	imes \left(\frac{k+5}{h-2}ight) = -1$$(k-7)(k+5) = -(h+2)(h-2)$$k^2 - 2k - 35 = -(h^2 - 4) = -h^2 + 4$$h^2 + k^2 - 2k - 39 = 0$. Thus,$E_1: x^2 + y^2 - 2y - 39 = 0$.For $E_2$,let a chord of $E_1$ have midpoint $(h, k)$. The equation of the chord is $T = S_1$,i.e.,$xh + yk - (y + k) - 39 = h^2 + k^2 - 2k - 39$.Since it passes through $R(1, 1)$,$h + k - (1 + k) - 39 = h^2 + k^2 - 2k - 39 \Rightarrow h^2 + k^2 - h - 2k + 1 = 0$.Checking options: $(A)$ $(-2, 7)$ gives $4 + 49 - 14 - 39 = 0$,but $P$ is the point of tangency,not $M$. $(D)$ $(0, 3/2)$ gives $0 + 9/4 - 3 - 39 
eq 0$,so it does not lie in $E_1$. Thus $(D)$ is true.

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