A) \[2ax+2by+({{a}^{2}}+{{b}^{2}}+4)=0\]
B) \[2\,ax\,+2by-({{a}^{2}}+{{b}^{2}}\,+4)=0\]
C) \[2ax-2by+({{a}^{2}}+{{b}^{2}}+4)=0\]
D) \[2ax-2by-({{a}^{2}}+{{b}^{2}}+4)=0\]
Correct Answer: B
Solution :
If two circles \[{{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0\]and \[{{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0\]cut orthogonally, then \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\] Let the equation of circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] It cuts the circle\[{{x}^{2}}+{{y}^{2}}=4\]orthogonally, if \[2g.0+2f.0=c-4\] \[\Rightarrow \] \[c=4\] \[\therefore \]Equation of circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+4=0\] \[\because \]It passes through the point (a, b). \[\therefore \] \[{{a}^{2}}+{{b}^{2}}+2ag+2fb+4=0\] Hence, locus of centre\[(-\text{ }g,-f)\]will be \[{{a}^{2}}+{{b}^{2}}-2xa-2yb+4=0\] \[\Rightarrow \]\[2ax+2by-({{a}^{2}}+{{b}^{2}}+4)=0\] Alternate Solution Let the centre of required circle be\[(-g,-f)\]. This circle cuts the circle\[{{x}^{2}}+{{y}^{2}}=4\]orthogonally. The centre and radius of circle\[{{x}^{2}}+{{y}^{2}}=4\]are (0, 0) and 2, respectively. \[\therefore \]\[{{g}^{2}}+{{f}^{2}}=4+{{(a+g)}^{2}}+{{(b+f)}^{2}}\] \[\Rightarrow \]\[{{g}^{2}}+{{f}^{2}}=4+{{a}^{2}}+{{g}^{2}}+2ag+2bf+{{b}^{2}}+{{f}^{2}}\] \[\Rightarrow \]\[4+{{a}^{2}}+{{b}^{2}}+2ag+2bf=0\] Hence, the locus of centre is \[2ax+2by-({{a}^{2}}+{{b}^{2}}+4)=0\]
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