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Question

If a circles passes through the point $$\left(a,b\right)$$ and cuts the circle $$x^{2}+y^{2}=4$$ orthogonally, then the locus of its centre is


A
2ax2by+(a2+b2+4)=0
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B
2ax+2by(a2+b2+4)=0
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C
2ax+2by+(a2+b2+4)=0
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D
2ax2ab(a2+b2+4)=0
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Solution

The correct option is B $$2ax+2by-\left(a^{2}+b^{2}+4\right)=0$$
Let the variables of the two circles is
$${x^2} + {y^2} + 2gx + 2fy + c = 0 \to (i)$$

If passes through $$(x,y)$$
if passes through $$(a, b)$$

Therefore,
$$\begin{array}{l} { a^{ 2 } }+{ b^{ 2 } }+2ga+2fb+c=0\,\,\,\,........ (ii) \\ Let\, \, { x^{ 2 } }+{ y^{ 2 } }=4\, \, orthogonally \\ 2\left( { g\times 0+f\times 0 } \right) =c-4 \\ c=4 \\ from\, \, equation\, \, \to (ii) \\ { a^{ 2 } }+{ b^{ 2 } }+2ga+2fb+4=0 \\ locus\, \, of\, \, centre(-g,-f)\, \, is \\ a{ q^{ 2 } }+{ b^{ 2 } }-2ax-2by+4=0 \\ 2ax+2by={ a^{ 2 } }+{ b^{ 2 } }+4 \\ 2ax+2by-\left( { { a^{ 2 } }+{ b^{ 2 } }+4 } \right)=0 \end{array}$$

Hence, this is the answer.

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