what are the coordinates of the center and the length of the radius of the circle?

Question

The equation of a circle is x^2+y^2-2x+6y+3=0. What are the coordinates of the center and the length of the radius of the circle.

Answer 1

Hi Ava, to find the coordinates of the circle and the radius, use the completing the square method.

 

x² + y² - 2x + 6y + 3 = 0

 

Rewrite:

 

x² - 2x + y² + 6y = -3

 

Complete square for x:

 

x² - 2x + 1 - 1

 

(x - 1)² -1

 

Complete square for y:

 

y² + 6y + 9 - 9

 

(y + 3)² - 9

 

Combine results:

 

(x - 1)² + (y + 3)² - 10 = -3

 

(x - 1)² + (y + 3)² = 7

 

Coordinates of center of circle:

 

(x - h)² + (y - k)² = r²

 

Where (h, k) is the location of the coordinates of the center of the circle and r is the radius.

 

Here, h = 1, k = -3, and r = √7

 

(h, k) = (1, -3)

 

Radius = √7

 

Questions?

Answer 2

Hi Ava;

 

We will need to complete the two squares in

order to get the format desired.

 

x² + y² - 2x + 6y + 3 = 0

 

We want:  (x-h)²+(y-k)² = r²

Center (h,k), radius r

 

(x² - 2x)+(y²+6y) + 3 = 0

Complete the squares

 

(x²-2x)

(1/2)(-2)=-1..., (-1)² = 1

Don't forget: If we add 1, we must also

subtract 1 to maintain equality.

(x²-2x+1)-1 = (x-1)² - 1

 

(y²+6y)

(1/2)6 = 3 ... 3²=9

Here we add 9, thus we also subtract 9.

(y²+6y+9)-9 = (y+3)² -9

 

(x-1)² - 1 + (y+3)² - 9 + 3 = 0

(x-1)² + (y+3)² - 7 = 0

(x-1)² + (y+3)² = 7

 

Center: (h,k) = (1, -3)

Radius: r = √7

 

Questions?