Given circles {x^2} + {y^2} - 4x - 5 = 0 and {x^2} + {y^2} + 6x - 2y + 6 = 0 . Let P be a point (α

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10-1.Circle and System of Circles

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Given the circles ${x^2} + {y^2} - 4x - 5 = 0$and ${x^2} + {y^2} + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha ,\beta )$such that the tangents from P to both the circles are equal, then

A

$2\alpha + 10\beta + 11 = 0$

B

$2\alpha - 10\beta + 11 = 0$

C

$10\alpha - 2\beta + 11 = 0$

D

$10\alpha + 2\beta + 11 = 0$

Solution

(c) Accordingly, ${\alpha ^2} + {\beta ^2} – 4\alpha – 5 $

$= {\alpha ^2} + {\beta ^2} + 6\alpha – 2\beta + 6$

$ \Rightarrow 10\alpha – 2\beta + 11 = 0$.

Standard 11

Mathematics

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