question_answer

The equations of perpendicular bisectors of two sides AB and AC of a triangle ABC are \[x+y+1=0\]and \[x-y+1=0.\]respectively. If circumradius of \[\Delta \,ABC\] is 2 units and the locus of vertex A is \[{{x}^{2}}+{{y}^{2}}+gx+c=0,\]then \[({{g}^{2}}+{{c}^{2}}),\] is equal to

A)  4                                

B)  5

C)  9                                

D)  13

Correct Answer: D

Solution :

The given lines intersect at (-1, 0) The vertex A lies on circle having centre at (-1, 0) and radius 2 units the vertex A lies on circle having centre at (-1, 0) and radius 2 units \[\Rightarrow \] Locus of A is \[{{(x+1)}^{2}}+{{y}^{2}}=4\] Hence g = 2 and c = -3  \[\Rightarrow \,\,{{g}^{2}}+{{c}^{2}}=13\]