Please help!!!!! MEDAL AND FAN!!!!! Give the center coordinates (x,y) of the circle:

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Please help!!!!! MEDAL AND FAN!!!!! Give the center coordinates (x,y) of the circle:

Mathematics
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\[(x+3)^{2}+(y-4)^{2}=36\]
equation of a circle (x-h)^2 + (y-k)^2 = r^2 center (h,k) and radius r
Center = (-3,4) This is because, as triciaal said, in the equation (x+3)2+(y−4)2=36, 3=h and 4=k (H,K) are the coordinates of the center.

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Other answers:

Oh ok
The definition of the circle is: the loci of a point which is moving at constant distance from a fixed point. So if the point is (h, k) and the constant distance is r, by the distance formula: \[\sqrt{(x-h)^2+(y-k)^2}=r\] After squaring, equation of the circle: \[(x-h)^2+(y-k)^2=r^2\]
When multiplied out, we obtain the "general form" of the equation of a circle. Notice that in this form we can clearly see that the equation of a circle has both x2 and y2 terms and these terms have the same coefficient (usually 1).
Wait so which explanation should I go with because I a bit confused
all are correct! different amount of details.
ok thx
yeah there all the same
But..... the first time I did this problem, I put -3,4 for my answer and got it wrong
ordered pairs should be in the form (h,k) that is you should have (-3,4)
Maybe they already included the ( ). Don't know.
oh ok I will put it in parentheses
Could y'all help me on on more problem?
*one
Sure!
\[(x + 1)^{2} + (y - 6)^{2}= 2.25\]
Would the coords be (-1,6)?
yes the center would be (-1,6)
Absolutly right!
thx

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