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

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Question

(b) Let point of contact be $P({x_1},\;{y_1})$.

This point lies on line

${x_1} + 2{y_1} = - 12$&hellip;.$(i)$

Gradient of $OP = {m_1} = \frac

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Let point of contact be $P({x_1},\;{y_1})$.

This point lies on line

${x_1} + 2{y_1} = - 12$&hellip;.$(i)$

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,

$	herefore \;{m_1}{m_2} = - 1$

$ \Rightarrow \left( {\frac{{{y_1} - 1}}{{{x_1} + 1}}} ight){m{ }}\left( {\frac{{ - 1}}{2}} ight) = - 1 \Rightarrow {y_1} - 1 = 2{x_1} + 2$

$ \Rightarrow 2{x_1} - {y_1} = - 3$&hellip;.$(ii)$

On solving equation $(i)$ and $(ii)$, we get

$({x_1},\;{y_1}) = \left( {\frac{{ - 18}}{5},\;\frac{{ - 21}}{5}} ight)$ .

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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