If the lengths of tangents drawn from a point | vedclass

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(C) Let $P(x_1, y_1)$ be the point. The length of the tangent from $P$ to a circle $S=0$ is $\sqrt{S}$.For the given circles,we normalize them to the

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $P(x_1, y_1)$ be the point. The length of the tangent from $P$ to a circle $S=0$ is $\sqrt{S}$.For the given circles,we normalize them to the form $x^2+y^2+2gx+2fy+c=0$:$S_1: x^2+y^2-8x+40=0$$S_2: x^2+y^2-5x+16=0$$S_3: x^2+y^2-8x+16y+160=0$Since the lengths of the tangents are equal,$S_1 = S_2 = S_3$ at point $P(x_1, y_1)$:$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-5x_1+16 = x_1^2+y_1^2-8x_1+16y_1+160$Equating $S_1 = S_3$:$-8x_1+40 = -8x_1+16y_1+160$ $\Rightarrow 16y_1 = -120$ $\Rightarrow y_1 = -\frac{15}{2}$Equating $S_1 = S_2$:$-8x_1+40 = -5x_1+16$ $\Rightarrow 3x_1 = 24$ $\Rightarrow x_1 = 8$Thus,the point $P$ is $\left(8, -\frac{15}{2}\right)$.

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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