A) \[{{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}}=a{{c}_{1}}+{{b}_{2}}\]
B) \[2({{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}})=b{{c}_{1}}+a{{c}_{2}}\]
C) \[b{{g}_{1}}{{g}_{2}}+a{{f}_{1}}{{f}_{2}}=b{{c}_{1}}+a{{c}_{2}}\]
D) \[{{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\]
Correct Answer: B
Solution :
Given equation of circles are \[a{{x}^{2}}+a{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0\] and \[b{{x}^{2}}+b{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0\] Or it can be rewritten as \[{{x}^{2}}+{{y}^{2}}+\frac{2{{g}_{1}}}{a}x+\frac{2{{f}_{1}}}{a}y+\frac{{{c}_{1}}}{a}=0\] and \[{{x}^{2}}+{{y}^{2}}+\frac{2{{g}_{2}}}{b}x+\frac{2{{f}_{2}}}{b}y+\frac{{{c}_{2}}}{b}=0\] \[\therefore \] Centres of circles are \[\left( -\frac{{{g}_{1}}}{a},-\frac{{{f}_{1}}}{a} \right)\]and \[\left( -\frac{{{g}_{2}}}{b},-\frac{{{f}_{2}}}{b} \right)\] respectively. We know, if two circles cut orthogonally, then \[2({{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}})={{c}_{1}}+{{c}_{2}}\] \[\Rightarrow \] \[2\left( \frac{{{g}_{1}}{{g}_{2}}}{ab}+\frac{{{f}_{1}}{{f}_{2}}}{ab} \right)=\frac{{{c}_{1}}}{a}+\frac{{{c}_{2}}}{b}\] \[\Rightarrow \] \[2({{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}})=b{{c}_{1}}+a{{c}_{2}}\]
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