[math]x^2+y^2=r^2[/math]

[math]x^2+y^2=13[/math][br](The radius is [math]CP=\sqrt{\left(x_C-x_P\right)^2+\left(y_C-y_P\right)^2}=\sqrt{13}[/math])

[math]x^2+y^2+6x-2y-31=0[/math] or [math]\left(x+3\right)^2+\left(y-1\right)^2=41[/math][br][br]One way to solve it is by considering that if [i]AB[/i] is the diameter, the midpoint of the segment [i]AB[/i] is the center [i]C[/i] of the circle.[br][i]CA[/i] (or [i]CB[/i], or 1/2[i] AB[/i]) is the radius. Plug the values of center and radius in the equation of the circle given center and radius, and expand the equation.

The graph of this equation is just a point, and precisely point C=(-5,-6).[br]Consider the general equation of a circle, [math]x^2+y^2+ax+by+c=0[/math]. In our case [math]a=10[/math], [math]b=12[/math] and [math]c=61[/math].[br]The equation represents a circle is and only if it has a center and a radius.[br][math]C=\left(-\frac{a}{2},-\frac{b}{2}\right)=\left(5,-6\right)[/math] and [math]r=\sqrt{\left(-\frac{a}{2}\right)^2+\left(-\frac{b}{2}\right)^2-c}=\sqrt{25+36-61}=0[/math][br]Being the radius 0, the circle is degenerated into its center [i]C[/i].