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  • Tangents PQ,PR are drawn to the circle  from a given po

Tangents PQ,PR are drawn to the circle \mathrm{c^2+y^2=a^2} from a given point \mathrm{P\left(x_1, y_1\right)} . The equation of the circum -circle of the triangle PQR is 

Option: 1

\mathrm{ x^2+y^2-x x_1+y y_1=0 }


Option: 2

\mathrm{ x^2+y^2-x x_1+2 y y_1=0 }


Option: 3

\mathrm{x^2+y^2+x x_1-2 y y_1=0}


Option: 4

\mathrm{ x^2+y^2-x x_1-y y_1=0 }


Answers (1)

best_answer

P be \mathrm{\left(x_1, y_1\right)} so that QR is chord of contact whose
equation is \mathrm{x x_1+y y_1-a^2=0} 0. The circle PQR thus passes through the intersection of given circle and chord QR and hence by \mathrm{S-\lambda\: P=0} its equation is

\mathrm{\left(x^2+y^2-a^2\right)-\lambda\left(x x_1+y y_1-a^2\right)=0} -------(i)

As it passes through \mathrm{P\left(x_1, y_1\right)}

\mathrm{\therefore\left(x_1^2+y_1^2-a^2\right)-\lambda\left(x_1^2+y_1^2-a^2\right)=0 \therefore \lambda=1}

Putting the value of \mathrm{\lambda} in (i) the required circle is \mathrm{x^2+y^2-x x_1-y y_1=0}  which is clearly passes through the centre (0, 0) of the given circle.

Posted by

Rakesh

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