Answer:
Centre: (2,1)
Radius:2^2
Explanation:
Circle radius given x^2 - 4x + y^2- 2y - 3 =0 r = 2^2
x^2 - 4x + y^2- 2y - 3 =0
Circle Equation: (x-a)^2+ (y-b)^2=r^2 is the circle equation with a radius, r centered at (a, b)
x^2 - 4x + y^2- 2y - 3 =0 in the form of the standard circle equation
(x-2)^2+(y-1)^2=(2,2^2)^2
Move the loose number to the right side
x^2 - 4x + y^2- 2y - 3 =0
Group x - variables and y - variables together
(x^2 - 4x) + (y^2 -2y) = 3
Convert x to square form
(x^2 - 4x + 4) + (y^2 - 2y) = 3 + 4
Convert to square form
(x -2)^2 + (y^2 - 2y) = 3 + 4
Convert y to square form
(x -2)^2 + (y^2 - 2y + 1) = 3 + 4 + 1
Convert to square form
(x-2)^2 + (y - 1)^2 = 3 + 4 + 1
Refine 3 + 4 + 1
(x-2)^2+(y-1)^2=8
Rewrite the standard form
(x-2)^2 + (y - 1)^2 = (2^2)^2
Therefore the circle properties are
(a, b) = (2,1), r = 2^2
And the radius is
2^2
CENTRE:
Move the loose number to the right
X^2-4x+y^2-2y = 3
Group x- variables and y- variables together
(x^2 - 4x) + (y^2 - 2y) = 3
Convert x to square form
(x^2 - 4x + 4) + (y^2 - 2y) = 3 + 4
Convert to square form
(x -2)^2 + (y^2 - 2y) = 3 + 4
Convert y to square form
(x -2)^2 + (y^2 - 2y + 1) = 3 + 4 + 1
Convert to square form
(x -2)^2 + (y - 1) = 3 + 4
Refine 3 + 4 + 1
(x- 2)^2 + (y - 1)^2 = 8
Rewrite in standard form
(x-2)^2 + (y - 1)^2 = (2^2)^2
Therefore the circle properties are:
(a, b) = (2, 1), r+2^2
And the center is:
(2, 1)