If straight line y = mx + c touches circle {x^2} + {y^2} - 2x - 4y + 3 = 0 at point (2, 3) , then c

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10-1.Circle and System of Circles

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If the straight line $y = mx + c$ touches the circle ${x^2} + {y^2} - 2x - 4y + 3 = 0$ at the point $(2, 3)$, then $c =$

A

$-3$

B

$4$

C

$5$

D

$-2$

Solution

(c) $y = mx + c$ is a tangent at $(2, 3)$.

Equation of tangent at $(2, 3) $ is $2x + 3y – (x + 2) – 2(y + 3) + 3 = 0$

$ \Rightarrow 2x + 3y – x – 2y – 2 – 6 + 3 = 0$

$ \Rightarrow x + y – 5 = 0$

$\Rightarrow c = 5$.

Standard 11

Mathematics

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(IIT-1975)