Find an equation of the circle that has center (-3,2) and passes through (4,-6) .

Question

anonymous · Answer

(x -h)^2 + (y - k)^2 =r^2
^equation of a circle. plug things in.

anonymous · Answer

(h,k) being the center

azureilai · Answer

equation of a circle is:   $$(x-h)^{2}+(y-k)^{2}=r^{2}$$
(h,k) is the center
substitute (x,y) to find r
rewrite equation

anonymous · Answer

i do that and come to sort(113)

anonymous · Answer

sqrt*

anonymous · Answer

r=sqrt113?

azureilai · Answer

yep

anonymous · Answer

and my equation would be (x--3)+(y-2)=113

azureilai · Answer

Don't forget the squared part in the left side (Refer to the standard equation I wrote above) you can leave 113 as it is though since it is already squared.

anonymous · Answer

i see, thanks team

anonymous · Answer

(4+3)^(2)+(-6-2)^(2)=r^2

49+64=r^2

113=r^2

equation would be

(x+3)^(2)+(y-2)^(2)=113

so yes you had it right:)

anonymous · Answer

woot thanks! <3