Answer:
(x-2)²+(y+7)² = 25
Explanation:
The general standard form of an equation is written in expressed as shown;
(x-a)²+(y-b)² = r²
(a,b) is the centre of the circle and r is the radius
Given the equation of the circle x^2 + y^2 - 4x + 14y +28 =0, we need to find the centre of the circle and its radius.
comparing the equation given with x²+y²+2gx+2fy+c =0
(-g, -f) is the centre
radius = √g²+f²-c
On comparison,
2gx = - 4x
g = -4/2
g = -2
2fy = 14y
f = 14/2
f = 7
c = 28
r = √(-2)²+7²-28
r = √4+49-28
r = √25
r = 5
comparing centre (a,b) to (-g, -f)
a = -g = -(-2)
a = 2
b = -f = -7
b = -7
substituting the centre (a, b) and the radius r into the equation
(x-a)²+(y-b)² = r²
The required equation will be (x-2)²+(y+7)² = 5²
(x-2)²+(y+7)² = 25