#### 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 Option: 2 Option: 3 Option: 4

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.

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