Question

find the center and radius of the circle with equation x^2 + y^2=2x-2y+7.

Answer 1

Given circleequation is x^2+y^2 = 2x-2y+7

Bring all terms to one side.

x^2+y^2-2x+2y-7 = 0

Compare it to the circle equation is x^2+y^2+2gx+2fy+c = 0.

2g = -2 then g = -1

2f = 2 then f = 1

Center (-g,-f) = (1,-1)

r = √(g^2+f^2-c)

r = √(1+1+7)

r = √9

Radius of given circle is 3.

Answer 2

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.

The equation is x² + y² = 2x - 2y + 7.

Rewrite the equation as x² - 2x + y² + 2y = 7.

Write the equation in standard form of a circle.

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

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

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

Add 1 and 1 to each side.

x² - 2x + 1 + y² + 2y + 1 = 7 + 1 + 1

(x - 1)² + (y + 1)² = 9

(x - 1)² + (y - (- 1))² = 3²

Compare the equation with standard form of a circle equation.

The center (h, k) is (1, - 1), and

The radius (r) is 3 units.