Equation of tangent to circle {x^2} + {y^2} = {a^2} which is perpendicular to straight line y = mx +

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10-1.Circle and System of Circles

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Equation of the tangent to the circle ${x^2} + {y^2} = {a^2}$ which is perpendicular to the straight line $y = mx + c$ is

A

$y = - \frac{x}{m} \pm a\sqrt {1 + {m^2}} $

B

$x + my = \pm {\rm{ }}a{\rm{ }}\sqrt {1 + {m^2}} $

C

$x + my = \pm a\sqrt {1 + {{(1/m)}^2}} $

D

$x - my = \pm a\sqrt {1 + {m^2}} $

Solution

(b) Line perpendicular to $y = mx + c$ is $y = – \frac{1}{m}x + \lambda $

and $m\lambda = \pm a\sqrt {1 + {m^2}} $

Hence required tangent is $my + x = \pm a\sqrt {1 + {m^2}} $.

Standard 11

Mathematics

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