question_answer

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 [MP PET 1998]

A)            \[\left( \frac{-7}{2},-4 \right)\]

B)            \[\left( \frac{-18}{5},\frac{-21}{5} \right)\]

C)            (2,-7)                                        

D)            (-2, -5)

Correct Answer: B

Solution :

           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,                    \[\therefore \ {{m}_{1}}{{m}_{2}}=-1\]                    \[\Rightarrow \left( \frac{{{y}_{1}}-1}{{{x}_{1}}+1} \right)\text{ }\left( \frac{-1}{2} \right)=-1\Rightarrow {{y}_{1}}-1=2{{x}_{1}}+2\]                    \[\Rightarrow 2{{x}_{1}}-{{y}_{1}}=-3\]                                                               ?. (ii)                    On solving equation (i) and (ii), we get                    \[({{x}_{1}},\ {{y}_{1}})=\left( \frac{-18}{5},\ \frac{-21}{5} \right)\] .