Question

I need to know how to find the radius and center of x^2 - 2x + y^2 – 16y = – 1.

Answer 1

Given circle equation x ^2 - 2x + y ^2 - 16y = -1

Add 1 to each side.

x ^2 + y ^2 - 2x - 16y + 1  = 0

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

c = 1

2g = -2

Divide each side by 2.

g  = -1

2f  = -16

Divide each side by 2.

f  = -8

Center (-g, -f ) = ( 1 , 8)

Radius(r ) = √(g ^2 + f ^2 - c )

= √(1 + 64 - 1)

= √ 64

= 8

Given circle center = ( 1 , 8 )

r  = 8.

 

Answer 2

The equation is x² - 2x + y² - 16y = - 1.

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 = - 16, so, (half the y coefficient)² = (- 16/2)²= 64.

Add 1 and 64 to each side.

x² - 2x + 1 + y² - 16y + 64 = - 1 + 1 + 64

(x - 1)² + (y - 8)² = 64

(x - 1)² + (y - 8)² = 8² .

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

The center (h, k) = (1, 8).

The radius (r) = 8 units.