# Determine the center and radius of the following circle:?

x^2+y^2+2x-14=0
Thanks!
asked Feb 7, 2013 in CALCULUS

x2 + y2 + 2x - 14 = 0

It can be written as x2 + 2(x)(1) + 12 - 12 + y2 + 2(y)(0) + 02  - 14 = 0

MATHEMATICAL FORMULAE: (a + b)2 = a2 + 2ab + b2

So, (x + 1)2 + (y - 0)2 - 1 - 14 = 0.

(x + 1)2 + (y - 0)2 - 15 = 0.

(x + 1)2 + (y - 0)2 = 15.

(x + 1)2 + (y - 0)2 = (√15)2.

Compare equation with standard from(x - h)2 + (y - k)2 = r2 and here center (h, k), radius is r.

So, center (-1, 0) and radius r = √15.

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 - 14 = 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 = 2, so, (half the x coefficient)² = (2/2)2= 1.

Here y coefficient = 0, so, (half the y coefficient)² = (0/2)2= 0.

Add 1 and 0 to each side.

x2 + 2x + 1 + y2 + 0 = 14 + 1 + 0

(x + 1)2 + (y + 0)2 = 15

(x - (- 1))2 + (y - (0))2 = (√15)2

Compare the equation with standard form of a circle equation.

The center is (- 1, 0) and