Find the center and radius of the circle x^2-14x+y^2+4y=-28

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The centre of the circlewith equation x^2+y^2+2gx+2fy+c =  0....(1)is  (-g , -f)

To  the centre of the circle whose equation is x^2-14x+y^2+4y=-28.

The given equation  could be brought to the standard form  (1) by just adding 28 to both sides and rewriting as:

x^2-14x+y^2+4y+28 =-28+28 = 0. Or

x^2-14x+y^2+4y + 28 = 0....(2).

Now  if (1) and (2) are to be the same equation, then term by term the are equal. So,

2gx = -14x and 2fy = 4y

Theferefore , -g = -14/-2 = 7.

-f = 4/-2 = -2.

Therefore (-g, -f) = (7,-2) are the coordinates of the centre of the equation of thecircle at (2).

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll have to re-write the equation in the standard form:

(x-h)^2 + (y-k)^2 = r^2

We'll have to complete the squares and we'll start with:

x^2 - 14x + b^2

We'll put x^2 = a^2 => x = a

-14x = -2*x*b => b = 14/2 => b = 7

b^2 = 49

To complete the square, we'll add and subtract 49:

x^2 - 14x + 49 - 49 = (x - 7)^2 - 49 (1)

y^2 + 4y + b^2

a^2 = y^2 => y = a

4y = 2*y*b => b = 4/2

b = 2 => b^2 = 4

To complete the square, we'll add and subtract 4:

y^2 + 4y + 4 - 4 = (y+2)^2 - 4 (2)

We'll add (1) + (2):

(x - 7)^2 - 49 + (y+2)^2 - 4 = -28

We'll move the constants to the right side:

(x - 7)^2 + (y+2)^2 = -28 + 4 + 49

(x - 7)^2 + (y+2)^2 = 25

The center of the circle is C(7 , -2) and the radius is r = 5.

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