Question

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

Express your answer in the form (a,b) without using spaces.


`x^2+y^2-2x+6y+9=0`

Answer 1

Explanation:

Rewrite this equation in standard form:

x² - 2x + 1 - 1 + y² + 6y + 9 = 0, or

(x - 1)² + (y + 3)² = 1

Compare this to:

(x + h)² + (y + 3)² = r²

We see here that (h, k), the center of the circle, is (1, -3), and the radius of the circle is 1.

Answer 2

the center is at (1, -3)

Explanation:

The center-radius form of the circle equation is in the format (x – h)2 + (y – k)2 = r2, with the center being at the point (h, k) and the radius being "r".

So we need to write the ecuation x^2 + y^2 - 2x + 6y + 9 = 0 in the format above.

So we have:

x^2 + y^2 - 2x + 6y + 9 = (x^2 -2x + 1) + (y^2 + 6y + 9) - 1

(x^2 -2x + 1) + (y^2 + 6y + 9) - 1 = (x-1)^2 + (y+3)^2 - 1

So now, looking at the equation: (x-1)^2 + (y+3)^2 = 1

We know that h=1 and k=-3. So the center is at (1, -3)