What is the center and the radius of the circle with equation (x+2)^2 + (y+3)^2 = 4
SOLUTION:
[tex](x + 2) {}^{2} + (y + 3) { }^{2} = 4[/tex]
[tex](x - a) {}^{2} + (y - b) {}^{2} = r {}^{2} [/tex]
IS THE CIRCLE EQUATION WITH A RADIUS R, CENTERED AT (A,B)
[tex](x + 2) {}^{2} + (y + 3) {}^{2} = 4[/tex]
IN THE FORM OF THE STANDARD CIRCLE EQUATION
[tex](x - ( - 2)) {}^{2} + (y - ( - 3)) {}^{2} =2 {}^{2} [/tex]
THEREFORE THE CIRCLE PROPERTIES ARE:
[tex](a,b)=(-2,-3),r=2[/tex]
ANSWER:
[tex]\large\orange{\boxed{circle, with, center at (-2,-3)}}[/tex]
[tex]\large\green{\boxed{radius \: \: R= 2}}[/tex]
(√^_^)/
