Point M moves along the circle (x - 4)^2 + (y | vedclass

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(D) Let the point of contact be $P(x_1, y_1)$. The equation of the tangent at $P$ is $(x_1 - 4)(x - 4) + (y_1 - 8)(y - 8) = 20$. Since this tangent pa

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$(B)$ and $(C)$ both

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(D) Let the point of contact be $P(x_1, y_1)$. The equation of the tangent at $P$ is $(x_1 - 4)(x - 4) + (y_1 - 8)(y - 8) = 20$. Since this tangent passes through $(-2, 0)$,we have $(x_1 - 4)(-2 - 4) + (y_1 - 8)(0 - 8) = 20$,which simplifies to $-6(x_1 - 4) - 8(y_1 - 8) = 20$. Dividing by $-2$,we get $3(x_1 - 4) + 4(y_1 - 8) = -10$,so $3x_1 - 12 + 4y_1 - 32 = -10$,or $3x_1 + 4y_1 = 34$. Also,$P(x_1, y_1)$ lies on the circle $(x_1 - 4)^2 + (y_1 - 8)^2 = 20$. Substituting $y_1 = \frac{34 - 3x_1}{4}$ into the circle equation: $(x_1 - 4)^2 + \left(\frac{34 - 3x_1}{4} - 8\right)^2 = 20$. $(x_1 - 4)^2 + \left(\frac{34 - 3x_1 - 32}{4}\right)^2 = 20 \implies (x_1 - 4)^2 + \left(\frac{2 - 3x_1}{4}\right)^2 = 20$. $16(x_1^2 - 8x_1 + 16) + (4 - 12x_1 + 9x_1^2) = 320$. $16x_1^2 - 128x_1 + 256 + 4 - 12x_1 + 9x_1^2 = 320$. $25x_1^2 - 140x_1 - 60 = 0 \implies 5x_1^2 - 28x_1 - 12 = 0$. $(5x_1 + 2)(x_1 - 6) = 0$. So,$x_1 = 6$ or $x_1 = -\frac{2}{5}$. If $x_1 = 6$,$y_1 = \frac{34 - 18}{4} = 4$. Point is $(6, 4)$. If $x_1 = -\frac{2}{5}$,$y_1 = \frac{34 - 3(-2/5)}{4} = \frac{34 + 6/5}{4} = \frac{176/5}{4} = \frac{44}{5}$. Point is $\left(-\frac{2}{5}, \frac{44}{5}\right)$. Thus,both $(B)$ and $(C)$ are correct.

Point $M$ moves along the circle $(x - 4)^2 + (y - 8)^2 = 20$. It then breaks away from the circle and moves along a tangent to the circle,passing through the point $(-2, 0)$ on the $x$-axis. The coordinates of the point on the circle at which the moving point broke away can be:

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