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10-1.Circle and System of Circles
medium
If the straight line $y = mx + c$ touches the circle ${x^2} + {y^2} - 4y = 0$, then the value of $c$ will be
A
$1 + \sqrt {1 + {m^2}} $
B
$1 - \sqrt {{m^2} + 1} $
C
$2(1 + \sqrt {1 + {m^2}} )$
D
$2 + \sqrt {1 + {m^2}} $
Solution
(c) Apply for tangency of line, centre being $(0, 2)$ and radius = $2$
$\left| {\frac{{ – 2 + c}}{{\sqrt {1 + {m^2}} }}} \right| = 2 $
$\Rightarrow {c^2} – 4c + 4 = 4 + 4{m^2}$
$ \Rightarrow c = \frac{{4 \pm \sqrt {16 + 16{m^2}} }}{2}$
or $c = 2 \pm 2\sqrt {1 + {m^2}} $
Most correct answer is $c = 2(1 + \sqrt {1 + {m^2}} )$.
Standard 11
Mathematics