If a circle,whose centre is (-1, 1),touches t | vedclass

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(B) Let the point of contact be $P(x_1, y_1)$.Since $P$ lies on the line $x + 2y + 12 = 0$,we have $x_1 + 2y_1 = -12$  $(i)$.The line $x + 2y + 12 = 0

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$\left( \frac{-18}{5}, \frac{-21}{5} \right)$

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(B) Let the point of contact be $P(x_1, y_1)$.Since $P$ lies on the line $x + 2y + 12 = 0$,we have $x_1 + 2y_1 = -12$  $(i)$.The line $x + 2y + 12 = 0$ has a slope $m_2 = -\frac{1}{2}$.The radius $OP$ is perpendicular to the tangent line,so the slope of $OP$ is $m_1 = -\frac{1}{m_2} = 2$.The slope of $OP$ is also given by $\frac{y_1 - 1}{x_1 + 1}$.Equating the slopes: $\frac{y_1 - 1}{x_1 + 1} = 2$ $\Rightarrow y_1 - 1 = 2x_1 + 2$ $\Rightarrow 2x_1 - y_1 = -3$  $(ii)$.Solving equations $(i)$ and $(ii)$:From $(ii)$,$y_1 = 2x_1 + 3$.Substituting into $(i)$: $x_1 + 2(2x_1 + 3) = -12$ $\Rightarrow x_1 + 4x_1 + 6 = -12$ $\Rightarrow 5x_1 = -18$ $\Rightarrow x_1 = -\frac{18}{5}$.Then $y_1 = 2(-\frac{18}{5}) + 3 = -\frac{36}{5} + \frac{15}{5} = -\frac{21}{5}$.Thus,the point of contact is $\left( -\frac{18}{5}, -\frac{21}{5} \right)$.

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

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