Power of Point with Respect to a Circle
Category : JEE Main & Advanced
Let \[P({{x}_{1}},{{y}_{1}})\] be a point outside the circle and PAB and PCD drawn two secants. The power of \[P({{x}_{1}},{{y}_{1}})\] with respect to \[S\equiv {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] is equal to PA. PB which is
\[x_{1}^{2}+y_{1}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c={{S}_{1}}\]
\[\therefore \,\,\,\,PA\,.\,PB={{(\sqrt{{{S}_{1}}})}^{2}}=\]
Square of the length of tangent.
If P is outside, inside or on the circle then PA. PB is \[+ve\], \[-ve\] or zero respectively.
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