A) PN.AB
B) \[2PN.AB\]
C) \[4PN.AB\]
D) None of these
Correct Answer: B
Solution :
| [b] Let the two given circles be |
| \[{{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+c=0\] ? (1) |
| And \[{{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+c=0\] ? (2) |
| Their centres are \[A(-{{g}_{1}},0)\] and \[B(-{{g}_{2}},0)\] |
| \[\therefore \,\,\,\,\,AB={{g}_{1}}-{{g}_{2}}\] |
| Let P be the point \[({{x}_{1}},{{y}_{1}}).\]Then, |
| \[PT=\sqrt{{{x}^{2}}_{1}+{{y}^{2}}_{1}+2{{g}_{1}}{{x}_{1}}+c};\] |
| \[PT=\sqrt{{{x}^{2}}_{1}+{{y}^{2}}_{1}+2{{g}_{2}}{{x}_{1}}+c}\] |
| Radical axis of (1) and (2) is \[2({{g}_{1}}-{{g}_{2}})x=0\] or \[x=0,\] |
| \[PN=\]Length of \[\bot \]from P on radical axis \[={{x}_{1}}.\] |
| \[\therefore \,\,\,\,P{{T}^{2}}-PT{{'}^{2}}\] |
| \[=({{x}^{2}}_{1}+{{y}^{2}}_{1}+2{{g}_{1}}{{x}_{1}}+c)-({{x}^{2}}_{1}+{{y}^{2}}_{1}+2{{g}_{2}}{{x}_{1}}+c)\] |
| \[=2{{x}_{1}}({{g}_{1}}-{{g}_{2}})=2PN.AB\] |
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