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10-1.Circle and System of Circles
hard
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
A
$\left( {2,\frac{5}{2}} \right)$
B
$\left( { - 2, - \frac{5}{2}} \right)$
C
$\left( { - 2,\frac{5}{2}} \right)$
D
$\left( {2, - \frac{5}{2}} \right)$
Solution
(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 $\left( { – 2,\; – \frac{5}{2}} \right)$.
Standard 11
Mathematics
