The equations of three circles are x^2 + y^2 | vedclass

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(D) The required point is the radical center of the three given circles.First,write the equations in the standard form $S = 0$:$S_1: x^2 + y^2 - 12x -

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None of these

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(D) The required point is the radical center of the three given circles.First,write the equations in the standard form $S = 0$:$S_1: x^2 + y^2 - 12x - 16y + 64 = 0$$S_2: x^2 + y^2 - 12x + 27 = 0$ (dividing by $3$)$S_3: x^2 + y^2 - 16x + 81 = 0$To find the radical center,solve the radical axes $S_1 - S_2 = 0$ and $S_2 - S_3 = 0$:$S_1 - S_2 = (-12x - 16y + 64) - (-12x + 27) = -16y + 37 = 0 \Rightarrow y = \frac{37}{16}$$S_2 - S_3 = (-12x + 27) - (-16x + 81) = 4x - 54 = 0 \Rightarrow x = \frac{54}{4} = \frac{27}{2}$The radical center is $(\frac{27}{2}, \frac{37}{16})$.Since this point is not among the given options,the correct answer is $(d)$.

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

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