Question

If centre of circles $x^2+y^2=25$ and $x^2+y^2-4x+9y+3=0$ are the endpoints of the diameter of a circle $S$, then equation of the circle $S$ is

• A. $x^2+y^2-2x+\frac{9y}{2}=0$
• B. $x^2+y^2+2x+\frac{9y}{2}=0$
• C. $x^2+y^2+2x-\frac{9y}{2}=0$
• D. $x^2+y^2+2x-4=0$

Answer 1

The correct option is A $x^2+y^2-2x+\frac{9y}{2}=0$

Circle$1:x^2+y^2=25$
Circle$2:x^2+y^2-4x+9y+3=0$
Center's of the circles are
$C_1\equiv (0,0),C_2\equiv (2,-\frac{9}{2})$

Since $C_1$ and $C_2$ are the endpoints of the diameter of circle $S$, so the equation of circle $S$ is
$(x-0)(x-2)+(y-0)(y+\frac{9}{2})=0\therefore x^2+y^2-2x+\frac{9y}{2}=0$