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Question: The equations of three circles are \(3 x ^ { 2 } + 3 y ^ { 2 } - 36 x + 81 = 0\) and \(x ^ { 2 } + ...

The equations of three circles are

3x2+3y236x+81=03 x ^ { 2 } + 3 y ^ { 2 } - 36 x + 81 = 0 and x2+y216x+81=0x ^ { 2 } + y ^ { 2 } - 16 x + 81 = 0The coordinates of the point from which the length of tangents drawn to each of the three circles is equal

A

(334,2)\left( \frac { 33 } { 4 } , 2 \right)

B

(2, 2)

C

(2,334)\left( 2 , \frac { 33 } { 4 } \right)

D

None

Answer

None

Explanation

Solution

The required point is the radical centre of the three given circles

Now, S1S2=016y+37=0S _ { 1 } - S _ { 2 } = 0 \Rightarrow - 16 y + 37 = 0,

S2S3=04x54=0S _ { 2 } - S _ { 3 } = 0 \Rightarrow 4 x - 54 = 0and S3S1=04x+16y+17=0S _ { 3 } - S _ { 1 } = 0 \Rightarrow - 4 x + 16 y + 17 = 0

Solving these equations, we get

.

Hence the required point is (272,3716)\left( \frac { 27 } { 2 } , \frac { 37 } { 16 } \right)