# A circle has the following equation. Find the center and radius. x^2+y^2=x/6-y/9-1/144

A circle has the following equation. Find the center and radius. x^2+y^2=x/6-y/9-1/144.

asked Feb 22, 2014 in GEOMETRY

Given circle equation is x^2+y^2 = x/6-y/9-1/144

Bring all terms to one side.

x^2+y^2-x/6+y/9+1/144 = 0

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

2g = -1/6 then g = -1/12

2f = 1/9 then f = 1/18

c = 1/144

Center (-g,-f) = (1/12,-1/18)

r = √(g^2+f^2-c)

r = √(1/144+1/324-1/144)

r = √1/324

r = 1/18

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 = x/6 - y/9 - 1/144.

Rewrite the equation as x2 - x/6 + y2 + y/9 = - 1/144.

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 = - 1/6, so, (half the x coefficient)² = (- 1/12)2= 1/144.

Here, y coefficient = 1/9, so, (half the y coefficient)² = (1/18)2= 1/324.

Add 1/144 and 1/324 to each side.

x2 - x/6 + 1/144 + y2 + y/9 + 1/324 = - 1/144 + 1/144 + 1/324

(x - 1/12)2 + (y + 1/18)2 = 1/324

(x - 1/12)2 + (y - (- 1/18))2 = (1/18)2

Compare the equation with standard form of a circle equation.

The center (h, k) is (1/12, - 1/18), and

The radius (r) is 1/18 units.