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If a circle, whose centre is $(-1, 1)$ touches the straight line $x + 2y + 12 = 0$, then the coordinates of the point of contact are
$\left( {\frac{{ - 7}}{2}, - 4} \right)$
$\left( {\frac{{ - 18}}{5},\frac{{ - 21}}{5}} \right)$
$(2,-7)$
$(-2, -5)$
Solution
(b) Let point of contact be $P({x_1},\;{y_1})$.
This point lies on line
${x_1} + 2{y_1} = – 12$….$(i)$
Gradient of $OP = {m_1} = \frac{{{y_1} – 1}}{{{x_1} + 1}}$
Gradient of $x + 2y + 12 = {m_2} = – \frac{1}{2}$
The two lines are perpendicular,
$\therefore \;{m_1}{m_2} = – 1$
$ \Rightarrow \left( {\frac{{{y_1} – 1}}{{{x_1} + 1}}} \right){\rm{ }}\left( {\frac{{ – 1}}{2}} \right) = – 1 \Rightarrow {y_1} – 1 = 2{x_1} + 2$
$ \Rightarrow 2{x_1} – {y_1} = – 3$….$(ii)$
On solving equation $(i)$ and $(ii)$, we get
$({x_1},\;{y_1}) = \left( {\frac{{ – 18}}{5},\;\frac{{ – 21}}{5}} \right)$ .
