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  • Two perpendicular tangents to the circle <img alt="\mathrm{x} 2+\mathrm{y} 2=\mathrm{a} 2" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%7D%202&plus;%5Cmathrm%7By%7D%202%3D

Two perpendicular tangents to the circle \mathrm{x} 2+\mathrm{y} 2=\mathrm{a} 2 meet at \mathrm{P}. Then the locus of \mathrm{P} has the equation

Option: 1

\mathrm{x}^{2}+\mathrm{y}^{2}=2 \mathrm{a}^{2}


Option: 2

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


Option: 3

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


Option: 4

None of these


Answers (1)

best_answer

The coordinates of \mathrm{P} be (\mathrm{h}, \mathrm{k}).Then the equation of the tangents drawn from \mathrm{P}(\mathrm{h}, \mathrm{k}) to \mathrm{x} 2+\mathrm{y} 2=\mathrm{a} 2 is
  \left(\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{a}^{2}\right)\left(\mathrm{h}^{2}+\mathrm{k}^{2}-\mathrm{a}^{2}\right)=\left(\mathrm{hx}+\mathrm{hy}-\mathrm{a}^{2}\right)^{2} \quad\left(\text { Using } S S^{\prime}=T^{2}\right)

This equation represents a pair of perpendicular lines.

Therefore, coefficient of \mathrm{x} 2+ coefficient of \mathrm{y} 2=0

\left(\mathrm{~h}^{2}+\mathrm{k}^{2}-\mathrm{a}^{2}-\mathrm{h}^{2}\right)+\left(\mathrm{h}^{2}+\mathrm{k}^{2}-\mathrm{a}^{2}-\mathrm{k}^{2}\right)=0
\mathrm{~h}^{2}+\mathrm{k}^{2}=2 \mathrm{a}^{2}

Hence, the locus of \mathrm{(h, k)} is \mathrm{x^{2}+y^{2}=2 a^{2}}.

Posted by

Ritika Jonwal

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