Question

What is the center and radius for the circle with equation 2x2 - 8x + 2y2 +12y + 14 = 0?

A A
(2,-3); r = 6
B
00
(293); r = 16
(-2,3); r = 6
0
D
(-2,3); r = 16

Answer 1

`(2,\, -3)`.

`r = √(6)`.

Explanation:

Let
`(a,\, b)` denote the center of this circle. Let
`r` (
`r > 0`) denote the radius of this circle. The equation of this circle would be:

`(x - a)^(2) + (y - b)^(2) = r^(2)`.

Expand to obtain an equivalent equation:

`x^(2) + (- 2\, a) + y^(2) + (- 2\, b) + (a^(2) + b^(2) - r^(2)) = 0`.

Since this equation and the given equation describe the same circle, their corresponding coefficients should match.

Notice that the coefficients of
`x^(2)` and
`y^(2)` in this equation are both
`1`. However, the corresponding coefficients in the given equation are both
`2`. Thus, divide both sides of the given equation by
`2\!` to match the coefficients of the
`x^(2)` and
`y^(2)` terms:

`x^(2) + (- 4\, x) + y^(2) + 6\, y + 7 = 0`.

• Coefficients of
  `x^(2)` and
  `y^(2)` are now matched after the division.
• Coefficients of
  `x` should match:
  `(-4) = (-2\, a)`, such that
  `a = 2`.
• Coefficients of
  `y` should match:
  `6 = (-2\, b)`, such that
  `b = (-3)`.

The constants should also match. Thus:

`a^(2) + b^(2) - r^(2) = 7`.

Substitute in
`a = 2` and
`b = (-3)`; given that
`r > 0`, the value of
`r` would be:

`r = \sqrt{2^(2) + 3^(2) - 7} = √(6)`.