Question

Find the center and radius of the circle given by the equation: x^2+y^2+8x+10y+37=0. The center is ? The radius is ?

Answer 1

\[x^{2}+8.x+y^{2}+10.y+37=0\]
equivalent
\[(x+4)^{2}-16 +(y+5)^{2}-25+32\]
\[(x+4)^{2}+(y+5)^{2}=9\]
C(-4,-5) and r=3

Answer 2

37 instead 32,so the radius is 2

Answer 3

when you are finding for the center of the radius and a circle. You need to identify first the equation if the center is at (h,k) or at (0,0). then you need to make them into standard form

General forms:
C (0,0) , x^2 + y^2 =r^2
C (h,k) , (x-h)^2 +(x-k)^2

Answer 4

based on your question, the center is at (h,k)

x^2+y^2+8x+10y+37=0

(x^2 + 8x) + (y^2 + 10y) = -37
(x^2 + 8x +16) +(y^2 +10y + 25) = -37 + 25 + 16
(x+4)^2 + (y+5)^2 = 4

C(-4,-5) r= 2
hope you understand it :)