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10-1.Circle and System of Circles
medium
The equations of the tangents to the circle ${x^2} + {y^2} - 6x + 4y = 12$ which are parallel to the straight line $4x + 3y + 5 = 0$, are
A
$3x - 4y - 19 = 0,\,\,3x - 4y + 31 = 0$
B
$4x + 3y - 19 = 0,\,\,4x + 3y + 31 = 0$
C
$4x + 3y + 19 = 0,\,\,4x + 3y - 31 = 0$
D
$3x - 4y + 19 = 0,3x - 4y + 31 = 0$
Solution
(c) Let equation of tangent be $4x + 3y + k = 0$, then $\sqrt {9 + 4 + 12} = \left| {\frac{{4(3) + 3( – 2) + k}}{{\sqrt {16 + 9} }}} \right|$
$ \Rightarrow 6 + k = \pm 25 \Rightarrow k = 19$ and $ – 31$.
Hence the tangents are $4x + 3y + 19 = 0$ and $4x + 3y – 31 = 0$.
Standard 11
Mathematics
