Question

What is the center of a circle whose equation is x^2+y^2-12x-2y+12=0

Answer 1

`(6,\, 1)`.

Explanation:

Let
`(a,\, b)` denote the center of this circle (where
`a` and
`b` are to be found.) If the radius of this circle is
`r`, the equation of this circle would be:

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

Expand this equation using binomial theorem:

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

Both this equation and the given equation
`x^(2) + y^(2) - 12\, x - 2\, y + 12 = 0` describe the same circle. Therefore, corresponding coefficients of the two equations should match one another:

• The coefficient of the
  `x` term should match in the two equations. Therefore:
  `(-2\, a) = (-12)`.
• The coefficient of the
  `y` term should also match in the two equations. Therefore:
  `(-2\, b) = (-2)`.

Solve these two equation for
`a` and
`b`:
`a = 6` and
`b = 1`.

Substitute
`a = 6` and
`b = 1` into
`(a,\, b)` to find the center of this circle:
`(6,\, 1)`.