Equation of a line through (7, 4) and touching circle, x^2 + y^2 - 6x + 4y - 3 = 0 is :

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10-1.Circle and System of Circles

normal

Equation of a line through $(7, 4)$ and touching the circle, $x^2 + y^2 - 6x + 4y - 3 = 0$ is :

A

$5x - 12y + 13 = 0$

B

$12x - 5y - 64 = 0$

C

$x - 7 = 0$

D

$(A)$ or $(C)$ both

Solution

$y – 4 = m (x – 7)$ ; using condition of tangency we get, $(3 – 2m)^2 = 4(1 -m^2)\, \Rightarrow \,m = 5/12$ or $/\infty$

Standard 11

Mathematics

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