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  • The pole of the line w.r.t. the circle <img alt="\mathrm{x^2+y^2

The pole of the line \mathrm{2 x+2 y-11=0} w.r.t. the circle \mathrm{x^2+y^2-4 x-6 y+12=0} is

Option: 1

(3,5)


Option: 2

(-3,-5)


Option: 3

(-4,-5)


Option: 4

(4,5)


Answers (1)

best_answer

Let \mathrm{\left(x_1, y_1\right)} be the pole of the line \mathrm{2 x+2 y-11=0} w.r.t. circle \mathrm{x^2+y^2-4 x-6 y+12=0}, then the polar of \mathrm{\left(x_1, y_1\right)} w.r.t. the circle be

\mathrm{ \begin{gathered} x x_1+y y_1-2\left(x+x_1\right)-3\left(y+y_1\right)+12=0 \\ \Rightarrow x\left(x_1-2\right)+y\left(y_1-3\right)+\left(12-2 x_1-3 y_1\right)=0 \end{gathered} }                 ...(i)

Now, line \mathrm{2 x+2 y-11=0} is same as (i).

\mathrm{ \begin{aligned} & \therefore \quad \frac{x_1-2}{2}=\frac{y_1-3}{2}=\frac{12-2 x_1-3 y_1}{-11} \\ & \Rightarrow 11 x_1-22=4 x_1+6 y_1-24 \text { and } 11 y_1-33=4 x_1+6 y_1-24 \\ & \Rightarrow 7 x_1-6 y_1+2=0 \text { and } 4 x_1-5 y_1+9=0 \end{aligned} }

On solving, we have, \mathrm{\left(x_1, y_1\right)=(4,5)}

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Pankaj Sanodiya

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