Question

Rewrite the equation in center - radius form. Decide whether or not the equation has a circle as its graph. If it does not, describe the graph. x^2 + y^2 + 8x - 10y + 41 = 0

Answer 1

The graph equation is given as;

$x^{2\text{ }}+y^2+8x-10y+41=0$

The format of the equation is ;

$(x-h)^2+(y-k)^2=r^2$

where ( h,k ) is the center and r is the radius

$x^2-8x+y^2-10y=-41$$x^2-8x+16+y^2-10y+25=\text{ -41 +16+25}$$(x-4)^2+(y-5)^2=0$

The equation has no circle as its graph because r=0