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(C) Let $P(x_1, y_1)$ be the point from which the tangents are drawn to the circles:$S_1 \equiv x^2+y^2-8x+40=0$$S_2 \equiv x^2+y^2-5x+16=0$ (Dividing

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$\left(8, -\frac{15}{2}\right)$

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(C) Let $P(x_1, y_1)$ be the point from which the tangents are drawn to the circles:$S_1 \equiv x^2+y^2-8x+40=0$$S_2 \equiv x^2+y^2-5x+16=0$ (Dividing by $5$)$S_3 \equiv x^2+y^2-8x+16y+160=0$Since the lengths of the tangents from $P$ to the circles $S_1, S_2, S_3$ are equal,we have $\sqrt{S_1} = \sqrt{S_2} = \sqrt{S_3}$,which implies $S_1 = S_2 = S_3$.Equating $S_1 = S_3$:$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-8x_1+16y_1+160$$40 = 16y_1+160$$16y_1 = -120 \Rightarrow y_1 = -\frac{120}{16} = -\frac{15}{2}$.Equating $S_1 = S_2$:$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-5x_1+16$$-8x_1+40 = -5x_1+16$$3x_1 = 24 \Rightarrow x_1 = 8$.Thus,the point $P$ is $\left(8, -\frac{15}{2}\right)$.

If the lengths of the tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the coordinates of point $P$ are:

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If the lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ is:

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