If tangent at left( {1,7} right) to curve {x^2} = y - 6 touches circle {x^2} + {y^2} + 16x + 1

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10-1.Circle and System of Circles

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If the tangent at $\left( {1,7} \right)$ to the curve ${x^2} = y - 6$ touches the circle ${x^2} + {y^2} + 16x + 12y + c = 0$ then the value of $c$ is:

A

$185$

B

$85$

C

$95$

D

$195$

(JEE MAIN-2018)

Solution

Equation of tangert at $(1,7)$ to ${x^2} = y – 6$ is

$2x – y + 5 = 0$

Now, perpentdicular from center $O(-8,-6)$to

$2x – y + 5 = 0$ should be equal to radium of the circle

$\left| {\frac{{ – 16 + 6 + 5}}{{\sqrt 5 }}} \right| = \sqrt {64 + 36 – C} $

$ \Rightarrow \sqrt 5 = \sqrt {100 – c} \Rightarrow c = 95$

Standard 11

Mathematics

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