If polar of a circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] with respect to \[(x',y')\] is \[Ax+By+C=0,\] then its pole will be

\[\left( \frac{{{a}^{2}}A}{C},\frac{{{a}^{2}}B}{C} \right)\]

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

\[\left( \frac{{{a}^{2}}C}{A},\frac{{{a}^{2}}C}{B} \right)\]

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

If polar of a circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] with respect to

1.	If polar of a circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] with respect to \[(x',y')\] is \[Ax+By+C=0,\] then its pole will be
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.	\[\left( \frac{{{a}^{2}}C}{-A},\frac{{{a}^{2}}C}{-B} \right)\]
Answer» B. \[\left( \frac{{{a}^{2}}A}{C},\frac{{{a}^{2}}B}{C} \right)\]

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If polar of a circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] with respect to \[(x',y')\] is \[Ax+By+C=0,\] then its pole will be

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