Question

Choose the equation for the given circle in standard form: x^2 + y^2 - 4x + 14y +28 =0

Answer 1

Answer:

(x-2)²+(y+7)² = 25

Step-by-step 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