The polar of the origin (0, 0) with respect t | vedclass

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(C) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1)

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${c^2} = {r^2}({\lambda ^2} + {\mu ^2})$

Vedclass · Answer

(C) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.For the point $(0, 0)$ and the circle $x^2 + y^2 + 2\lambda x + 2\mu y + c = 0$,the polar is $\lambda(x + 0) + \mu(y + 0) + c = 0$,which simplifies to $\lambda x + \mu y + c = 0$.This line touches the circle $x^2 + y^2 = r^2$ if the perpendicular distance from the center $(0, 0)$ to the line is equal to the radius $r$.The perpendicular distance $d$ from $(0, 0)$ to $\lambda x + \mu y + c = 0$ is given by $d = \frac{|c|}{\sqrt{\lambda^2 + \mu^2}}$.Setting $d = r$,we get $\frac{|c|}{\sqrt{\lambda^2 + \mu^2}} = r$.Squaring both sides,we obtain $\frac{c^2}{\lambda^2 + \mu^2} = r^2$,which implies $c^2 = r^2(\lambda^2 + \mu^2)$.

The polar of the origin $(0, 0)$ with respect to the circle ${x^2} + {y^2} + 2\lambda x + 2\mu y + c = 0$ touches the circle ${x^2} + {y^2} = {r^2}$,if

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