Determine the center and radius of the following circle:?

Question

x^2+y^2+2x-14=0
Thanks!

Answer 1

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.

Add 15 to each side.

(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.

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 + 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

The radius is √15 units.