Consider the circle S: x^2 + y^2 = 1 and poin | vedclass

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(C) The tangent to the circle $S: x^2 + y^2 = 1$ at $P(0, -1)$ is $y = -1$.The incident ray passes through $(-3, -1)$ and hits the tangent $y = -1$ at

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$3x - 4y + 5 = 0$

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(C) The tangent to the circle $S: x^2 + y^2 = 1$ at $P(0, -1)$ is $y = -1$.The incident ray passes through $(-3, -1)$ and hits the tangent $y = -1$ at $(-3, -1)$.Let the reflected ray be $y + 1 = m(x + 3)$,which simplifies to $mx - y + 3m - 1 = 0$.Since this reflected ray is also tangent to the circle $S: x^2 + y^2 = 1$,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $r = 1$.$\frac{|m(0) - (0) + 3m - 1|}{\sqrt{m^2 + (-1)^2}} = 1$$|3m - 1| = \sqrt{m^2 + 1}$Squaring both sides: $(3m - 1)^2 = m^2 + 1$$9m^2 - 6m + 1 = m^2 + 1$$8m^2 - 6m = 0$$2m(4m - 3) = 0$So,$m = 0$ or $m = \frac{3}{4}$.If $m = 0$,the line is $y = -1$,which is the tangent itself. The reflected ray is $m = \frac{3}{4}$.Substituting $m = \frac{3}{4}$ into the equation: $y + 1 = \frac{3}{4}(x + 3)$$4y + 4 = 3x + 9$$3x - 4y + 5 = 0$.

Consider the circle $S: x^2 + y^2 = 1$ and point $P(0, -1)$ on it. $A$ ray of light passes through the point $(-3, -1)$ and reflects from the tangent to $S$ at $P$. After reflection,it becomes tangent to the circle $S$. Find the equation of the reflected ray.

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