JEE MAIN - Mathematics (2005 - No. 46)
If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2} = {p^2}$$ orthogonally, then the equation of the locus of its centre is :
$${x^2}\, + \,{y^2} - \,3ax\, - \,4\,by\,\, + \,({a^2}\, + \,{b^2} - {p^2}) = 0$$
$$2ax\, + \,\,2\,by\,\, - \,({a^2}\, - \,{b^2} + {p^2}) = 0$$
$${x^2}\, + \,{y^2} - \,2ax\, - \,\,3\,by\,\, + \,({a^2}\, - \,{b^2} - {p^2}) = 0$$
$$2ax\, + \,\,2\,by\,\, - \,({a^2}\, + \,{b^2} + {p^2}) = 0$$
Explanation
Let the center be $$\left( {\alpha ,\beta } \right)$$
As It cuts the circle $${x^2} + {y^2} = {p^2}$$ orthogonally
$$\therefore$$ Using $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},\,\,$$ we get
$$2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0$$
$$ = {c_1} - {p^2} \Rightarrow {c_1} = {p^2}$$
Let equation of circle is
$${x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0$$
It passes through
$$\left( {a,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0$$
$$\therefore$$ Locus of $$\left( {\alpha ,\beta } \right)$$ is
$$\therefore$$ $$2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0.$$
As It cuts the circle $${x^2} + {y^2} = {p^2}$$ orthogonally
$$\therefore$$ Using $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},\,\,$$ we get
$$2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0$$
$$ = {c_1} - {p^2} \Rightarrow {c_1} = {p^2}$$
Let equation of circle is
$${x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0$$
It passes through
$$\left( {a,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0$$
$$\therefore$$ Locus of $$\left( {\alpha ,\beta } \right)$$ is
$$\therefore$$ $$2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0.$$
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