The equation of a circle is x2 + y2 + Cx + Dy + E = 0. If the radius of the circle is decreased without changing the coordinates of the center point, how are the coefficients C, D, and E affected?

Question

anonymous · Answer

any idea?

anonymous · Answer

I can see you are new here:) As a welcome gift, I give you my idea. hihihi
$$x^2+Cx+(\frac{C}{2})^2+y^2+Dy+(\frac{D}{2})^2-(\frac{C}{2})^2-(\frac{D}{2})^2 +E=0$$
$$(x+C/2)^2+(y+D/2)^2=\frac{C^2}{4}+\frac{D^2}{4}-E=\frac{C^2+D^2-4E}{4}$$
that is the standard formula of the equation center (-C/2,-D/2) and radius $\Huge r^2=\frac{C^2+D^2-4E}{4}$
C, D are unchangd, so that if r decreases, then E must increase .

jdoe0001 · Answer

@thaking     we're assuming you've covered the standard equation of a circle

so...  anything you may find confusing above there?

anonymous · Answer

E increases? Wtf? No.. thats NOT even one of the answers to chose from.. the fnck outta here with them false answers.. Tired of failing cause people put up wrong answers. -___-