question_answer

The locus of centre of a circle which passes through the origin and cuts off a length of 4 unit from the line\[x=3\]is

A) \[{{y}^{2}}+6x=0\]

B) \[{{y}^{2}}+6x=13\]

C) \[{{y}^{2}}+6x=10\]

D) \[{{x}^{2}}+6y=13\]

Correct Answer: B

Solution :

Let centre of circle be\[C(-g,\,\,-f)\], then equation of circle passing through origin be                 \[{{x}^{2}}+{{y}^{2}}+2gx+2fy=0\] \[\therefore \]Distance,\[d=|-g-3|\,\,=g+3\] In\[\Delta ABC,\,\,{{(BC)}^{2}}=A{{C}^{2}}+B{{A}^{2}}\] \[\Rightarrow \]               \[{{g}^{2}}+{{f}^{2}}={{(g+3)}^{2}}+{{2}^{2}}\] \[\Rightarrow \]               \[{{g}^{2}}+{{f}^{2}}={{g}^{2}}+6g+9+4\] \[\Rightarrow \]               \[{{f}^{2}}=6g+13\] Hence, required locus is\[{{y}^{2}}+6x=13\]