A) \[\left( \frac{{{a}^{2}}A}{-C},\frac{{{a}^{2}}B}{-C} \right)\]
B) \[\left( \frac{{{a}^{2}}A}{C},\frac{{{a}^{2}}B}{C} \right)\]
C) \[\left( \frac{{{a}^{2}}C}{A},\frac{{{a}^{2}}C}{B} \right)\]
D) \[\left( \frac{{{a}^{2}}C}{-A},\frac{{{a}^{2}}C}{-B} \right)\]
Correct Answer: A
Solution :
Polar of the circle is \[xx'+yy'={{a}^{2}}\], but it is given by \[Ax+By+C=0\], then \[\frac{x'}{A}=\frac{y'}{B}=\frac{{{a}^{2}}}{-C}\] Hence pole is\[\left( \frac{{{a}^{2}}A}{-C},\ \frac{{{a}^{2}}B}{-C} \right)\].
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