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(B) The point having the same power with respect to three circles is their radical center. To find the radical center,we find the equations of the rad

Vedclass · Accepted Answer

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

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(B) The point having the same power with respect to three circles is their radical center. To find the radical center,we find the equations of the radical axes by subtracting the circle equations. Let $S_1: x^2+y^2-8x+40=0$ $S_2: x^2+y^2-5x+16=0$ $S_3: x^2+y^2-8x+16y+160=0$ Radical axis $S_1 - S_2 = 0$: $(-8x+5x) + (40-16) = 0 \implies -3x + 24 = 0 \implies x = 8$. Radical axis $S_1 - S_3 = 0$: $(-8x+8x) - 16y + (40-160) = 0 \implies -16y - 120 = 0 \implies 16y = -120 \implies y = -\frac{120}{16} = -\frac{15}{2}$. The radical center is the intersection of these axes,which is $\left(8, -\frac{15}{2}\right)$.

The point which has the same power with respect to each of the circles $x^2+y^2-8x+40=0$,$x^2+y^2-5x+16=0$,and $x^2+y^2-8x+16y+160=0$ is

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