What is the equation of the circle with center (0,0) that passes through the point (5,-4)

Question

ashleyisakitty · Answer

I got (x+5)^2+(y+4)^2=41, however in my answer choices the only answer close to this is (x-(-4))^2+(y-4))^2=41.. or (x-5)^2+(y-(-4))^2=9. Where did I go wrong?

ashleyisakitty · Answer

@Agent47

agent47 · Answer

center is (0, 0), so you have:
x^2+y^2 = r^2, where r is the radius.

ashleyisakitty · Answer

I got 41 as the radius, and then used 6.4

agent47 · Answer

It passes through the point (5,-4), so:
r^2 = (0+5)^2+(0-4)^2 = 25+16

agent47 · Answer

so r^2 = 41.

agent47 · Answer

The equation is:
x^2+y^2 = 41

agent47 · Answer

Center is (0, 0), the equation of a circle is:
(x-h)^2 + (y-k)^2 = r^2 where (h, k) is the center.

ashleyisakitty · Answer

Hmm.. My steps were to find the radius and then go like
(x + 5)^2 + (y + 3)^2 = 6.4^2

(x + 5)^2 + (y + 3)^2 = 41

which got me to (x+5)^2+(y+4)^2=41..

agent47 · Answer

yea but you shouldn't bother taking the square root of the radius, since the equation wants it squared anyway.