What are the coordinates of the center of the circle shown below?

Question

What is the radius of the circle shown below?

x^2+y^2-2x+6y+9=0 

Answer 1

The standard form of the circle equation is ( x - h )² + ( y - k )² = r² , where, (h, k) is the center of the circle, and r is the radius.

Given circle equation is x² + y² - 2x + 6y + 9 = 0.

x² - 2x + y² + 6y + 9 = 0

First write the equation in standard form of a circle, then we can identify the center and radius of the circle.

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)² to each side of the expression.

Here, x coefficient = -2, so, (half the x coefficient)² = (-2/2)²= 1.

Here, y coefficient = 6, so, (half the y coefficient)² = (-6/2)²= 9.

Add 1 and 9 to each side.

x² - 2x + 1 + y² + 6y + 9 + 9 = 0 + 1 + 9

x² - 2(x)(1) + 1² + y² + 2(y)(3) + 3² + 9 = 10

(x - 1)² + (y + 3)² + 9 = 10

(x - 1)² + (y + 3)² = 10 - 9

(x - 1)² + (y + 3)² = 1

[ x - 1 ]² + [ y - (-3) ]²  = (1)² .

Compare the equation with standard form of a circle equation and write the coefficients.

h = 1 , k = -3 , r =1.

Solution :

The center (h, k) is (1, -3) and The radius (r) is 1 unit.