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10-1.Circle and System of Circles
hard
If the equation of one tangent to the circle with centre at $(2, -1)$ from the origin is $3x + y = 0$, then the equation of the other tangent through the origin is
A
$3x - y = 0$
B
$x + 3y = 0$
C
$x - 3y = 0$
D
$x + 2y = 0$
Solution
(c) Centre is $(2,\; – 1)$.
Therefore $r = \left| {\frac{{3(2) – 1}}{{\sqrt {10} }}} \right|\; = \frac{5}{{\sqrt {10} }}$
Now draw a perpendicular on $x – 3y = 0$, we get
$r = \left| {\frac{{2 – 3( – 1)}}{{\sqrt {10} }}} \right|\; $
$= \frac{5}{{\sqrt {10} }}$.
Standard 11
Mathematics
