A) \[2ax+2by+({{a}^{2}}+{{b}^{2}}+4)=0\]
B) \[2ax+2by-({{a}^{2}}+{{b}^{2}}+4)=0\]
C) \[2ax-2by+({{a}^{2}}+{{b}^{2}}+4)=0\]
D) \[2ax-2by-({{a}^{2}}+{{b}^{2}}+4)=0\]
Correct Answer: B
Solution :
Let the equation of circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] 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) \[\therefore \] \[{{a}^{2}}+{{b}^{2}}+2ag+2fb+4=0\] 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 is\[(-g,-f)\]. 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. \[\therefore \]\[{{g}^{2}}+{{f}^{2}}=4+{{(a+g)}^{2}}+{{(b+f)}^{2}}\] \[\Rightarrow \]\[{{g}^{2}}+{{f}^{2}}=4+{{a}^{2}}+{{g}^{2}}+2ag+2bf+{{b}^{2}}+{{f}^{2}}\] \[\Rightarrow \] \[4+{{a}^{2}}+{{b}^{2}}+2ag+2bf=0\] So the locus of centre is \[2ax+2by-({{a}^{2}}+{{b}^{2}}+4)=0\]
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