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  • The locus of the point whose polars w.r.t. the circle <img alt="\mathrm{x^2+y^2-4 x-4 y-8=0\, \, and \, \, x^2+y^2-2 x+6 y-2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5

The locus of the point whose polars w.r.t. the circle \mathrm{x^2+y^2-4 x-4 y-8=0\, \, and \, \, x^2+y^2-2 x+6 y-2=0} are mutually perpendicular, is

Option: 1

\mathrm{x^2+y^2-3 x+y=4}


Option: 2

\mathrm{x^2+y^2+3 x-y=4}


Option: 3

\mathrm{x^2+y^2+3 x+y=4}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

Let polar of the point \mathrm{ \left(x_1, y_1\right)}w.r.t. the given circles be mutually perpendicular.

\mathrm{ \begin{aligned} & \text { i.e., } x x_1+y y_1-2\left(x+x_1\right)-2\left(y+y_1\right)-8=0 \\\\ & \text { and } x x_1+y y_1-\left(x+x_1\right)+3\left(y+y_1\right)-2=0 \\\\ & \Rightarrow x\left(x_1-2\right)+y\left(y_1-2\right)-\left(2 x_1+2 y_1+8\right)=0 \\\\ & \text { and } x\left(x_1-1\right)+y\left(y_1+3\right)-\left(x_1-3 y_1+2\right)=0 \\\\ & \therefore m_1 m_2=-1 \\\\ & \Rightarrow(-1)^2\left(\frac{x_1-2}{y_1-2}\right)\left(\frac{x_1-1}{y_1+3}\right)=-1 \\\\ & \Rightarrow\left(x_1-2\right)\left(x_1-1\right)+\left(y_1-2\right)\left(y_1+3\right)=0 \\\\ & \Rightarrow x_1^2+y_1^2-3 x_1+y_1-4=0 \end{aligned} }

Thus, locus of \mathrm{ \left(x_1, y_1\right)\, \, is \, \, x^2+y^2-3 x+y=4}

Posted by

shivangi.bhatnagar

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