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 Find the equations of the tangents drawn from the point A(3, 2) to the circle \mathrm{x^2+y^2+4 x+6 y+8=0}

Option: 1

\mathrm{2 x-y-4=0 \: and\: x-2 y+1=0}


Option: 2

\mathrm{x-2 y-4=0 \: and\: 2 x+y-1=0}


Option: 3

\mathrm{2 x+y-4=0\: and\: x+2 y-1=0}


Option: 4

none of these


Answers (1)

best_answer

Joint equation of the pair of tangents drawn from A(3, 2) to the given circle\mathrm{x^2+y^2+4 y+6 y+8=0} can be written in the usual notation \mathrm{\mathrm{T}^2=\mathrm{SS}_1}namely,

\mathrm{\begin{aligned} & {[3 x+2 y+2(x+3)+3(y+2)+8]^2=\left[x^2+y^2+4 x+6 y+8\right][9+4+12+12+8]} \\ & \text { or }(5 x+5 y+20)^2=45\left(x^2+y^2+4 x+6 y+8\right) \\ & \text { or } 5\left(x^2+y^2+2 x y+8 x+8 y+16\right)=9\left(x^2+y^2+4 x+6 y+8\right) \\ & \text { or }\left(2 x^2+2 y^2-5 x y-2 x+7 y-4\right)=0 \\ & \text { or }(2 x-y-4)(x-2 y+1)=0 \end{aligned}}

Hence the required tangents to the circle from A(3, 2) are  2x – y – 4 = 0 and x – 2y + 1 = 0.

 

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shivangi.shekhar

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