what is the center and radius of a circle with the equation

Question

x^2 +y^2-4x+6y+1=0.

Answer 1

Given equation x ^2 + y ^2 - 4x  + 6y + 1 = 0

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

c  = 1

2g  = -4

g  = -2

2f  = 6

f  = 3

Center ( -g , -f ) = ( 2 , -3 )

Radius = √(g ^2 + f ^2 - c)

= √(4 + 9 - 1) = √12

Radius of circle(r) = 3.46.

Center = ( 2 , -3 )

r  = 3.46.

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 - 4x + 6y + 1 = 0.

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 = - 4, so, (half the x coefficient)² = (- 4/2)²= 4.

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

Add 4 and 9 to each side.

x² - 4x + 4 + y² + 6y + 9 + 1 = 0 + 4 + 9

(x - 2)² + (y + 3)² = 13 - 1 = 12

(x - 2)² + (y - (- 3))² = (2√3)² .

Compare the equation with standard form of a circle equation.

The center (h, k) is (2, - 3), and

The radius (r) is 2√3 units.