If the polar of a circle x^2 + y^2 = a^2 with | vedclass

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(A) The equation of the polar of the circle $x^2 + y^2 = a^2$ with respect to the point $(x', y')$ is given by $xx' + yy' = a^2$,which can be rewritte

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$\left( \frac{a^2 A}{-C}, \frac{a^2 B}{-C} \right)$

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(A) The equation of the polar of the circle $x^2 + y^2 = a^2$ with respect to the point $(x', y')$ is given by $xx' + yy' = a^2$,which can be rewritten as $x'x + y'y - a^2 = 0$.It is given that the equation of the polar is $Ax + By + C = 0$.Comparing the coefficients of the two equations,we have:$\frac{x'}{A} = \frac{y'}{B} = \frac{-a^2}{C}$From this,we get:$x' = \frac{-a^2 A}{C} = \frac{a^2 A}{-C}$$y' = \frac{-a^2 B}{C} = \frac{a^2 B}{-C}$Therefore,the pole $(x', y')$ is $\left( \frac{a^2 A}{-C}, \frac{a^2 B}{-C} \right)$.

If the polar of a circle $x^2 + y^2 = a^2$ with respect to a point $(x', y')$ is $Ax + By + C = 0$,then its pole will be:

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