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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