The co-ordinates of the point from where the | vedclass

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(b) Length of tangents is same $i.e.$, $\sqrt {{S_1}} = \sqrt {{S_2}} = \sqrt {{S_3}} $.

We get the point from where tangent is drawn, by solving t

Vedclass · Accepted Answer

$\left( { - 2, - \frac{5}{2}} \right)$

Vedclass · Answer

(b) Length of tangents is same $i.e.$, $\sqrt {{S_1}} = \sqrt {{S_2}} = \sqrt {{S_3}} $.

We get the point from where tangent is drawn, by solving the $3$ equations for $x$ and $y$.

$i.e.$, ${x^2} + {y^2} = 1$ ,

${x^2} + {y^2} + 8x + 15 = 0$ and ${x^2} + {y^2} + 10y + 24 = 0$

or $8x + 16 = 0$ and $10y + 25 = 0$

$ \Rightarrow x = - 2$ and $y = - \frac{5}{2}$

Hence the point is $\left( { - 2,\; - \frac{5}{2}} \right)$.

The co-ordinates of the point from where the tangents are drawn to the circles ${x^2} + {y^2} = 1$, ${x^2} + {y^2} + 8x + 15 = 0$ and ${x^2} + {y^2} + 10y + 24 = 0$ are of same length, are

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