Given this equation find the radius and center of a circle.

Question

x^2+6x+y^2-5y=-1/4.

Answer 1

Given circle equation x ^2 + 6x + y ^2 - 5y = -1/4

Add (1/4) to each side.

x ^2 + y ^2 + 6x - 5y + 1/4  = 0

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

c = 1/4

2g = 6

Divide each side by 2.

g  = 3

2f  = -5

Divide each side by 2.

f  = -5/2

Center (-g, -f ) = ( -3 , 5/2)

Radius(r ) = √(g ^2 + f ^2 - c )

= √[9 + (25/4) - (1/4)]

= √ 9 + (24/4)

= √ (9 + 6)

= √15

= 3.87

Given circle center = ( -3 , 5/2)

r  = 3.87.

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 + 6x + y² - 5y = - 1/4.

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

Here, y coefficient = - 5, so, (half the y coefficient)² = (- 5/2)²= 25/4.

Add 9 and 25/4 to each side.

x² + 6x + 9 + y² - 5y + 25/4 = - 1/4 + 9 + 25/4

(x + 3)² + (y - 5/2)² = 6 + 9 = 15

(x - (- 3))² + (y - 5/2))² = (√15)² .

Compare the equation with standard form of a circle equation.

The center (h, k) is (- 3, 5/2), and

The radius (r) is √15 units.