JEE MAIN - Mathematics (2004 - No. 44)
If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2} = 4$$ orthogonally, then the locus of its centre is :
$$2ax\, - 2by\, - ({a^2}\, + \,{b^2} + 4) = 0$$
$$2ax\, + 2by\, - ({a^2}\, + \,{b^2} + 4) = 0$$
$$2ax\, - 2by\, + ({a^2}\, + \,{b^2} + 4) = 0$$
$$2ax\, + 2by\, + ({a^2}\, + \,{b^2} + 4) = 0$$
Explanation
Let the variable circle is
$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,...\left( 1 \right)$$
It passes through $$(a,b)$$
$$\therefore$$ $${a^2} + {b^2} + 2ga + 2fb + c = 0\,\,\,\,\,\,\,...\left( 2 \right)$$
$$(1)$$ cuts $${x^2} + {y^2} = 4$$ orthogonally
$$\therefore$$ $$2\left( {g \times 0 + f \times 0} \right) = c - 4 \Rightarrow c = 4$$
$$\therefore$$ from $$(2)$$ $$\,\,\,{a^2} + {b^2} + 2ga + 2fb + 4 = 0$$
$$\therefore$$ Locus of center $$\left( { - g, - f} \right)$$ is
$${a^2} + {b^2} - 2ax - 2by + 4 = 0$$
or $$2ax + 2by = {a^2} + {b^2} + 4$$
$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,...\left( 1 \right)$$
It passes through $$(a,b)$$
$$\therefore$$ $${a^2} + {b^2} + 2ga + 2fb + c = 0\,\,\,\,\,\,\,...\left( 2 \right)$$
$$(1)$$ cuts $${x^2} + {y^2} = 4$$ orthogonally
$$\therefore$$ $$2\left( {g \times 0 + f \times 0} \right) = c - 4 \Rightarrow c = 4$$
$$\therefore$$ from $$(2)$$ $$\,\,\,{a^2} + {b^2} + 2ga + 2fb + 4 = 0$$
$$\therefore$$ Locus of center $$\left( { - g, - f} \right)$$ is
$${a^2} + {b^2} - 2ax - 2by + 4 = 0$$
or $$2ax + 2by = {a^2} + {b^2} + 4$$
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