Question

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

Answer 1

Answer:

$(6,\, 1)$.

Step-by-step 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: $(6,\, 1)$.