The equation of three circles are {x^2} + {y^ | vedclass

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Question

(d) The required point is the radical centre of the three given circles.

Now, ${S_1} - {S_2} = 0 \Rightarrow - 16y + 37 = 0$

${S_2} - {S_3} = 0\

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(d) The required point is the radical centre of the three given circles.

Now, ${S_1} - {S_2} = 0 \Rightarrow - 16y + 37 = 0$

${S_2} - {S_3} = 0\, \Rightarrow \,4x - 54 = 0$

${S_3} - {S_1} = 0\, \Rightarrow \, - 4x + 16y + 17 = 0$

Solving these equations, we get

$x = \frac{{54}}{4},\,\,y = \frac{{37}}{{16}}$$ \Rightarrow $$x = \frac{{27}}{2},\,y = \frac{{37}}{{16}}$

Hence the required point is $\left( {\frac{{27}}{2},\,\frac{{37}}{{16}}} \right)$.

The equation of three circles are ${x^2} + {y^2} - 12x - 16y + 64 = 0,$ $3{x^2} + 3{y^2} - 36x + 81 = 0$ and ${x^2} + {y^2} - 16x + 81 = 0.$ The co-ordinates of the point from which the length of tangent drawn to each of the three circle is equal is

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