The Equation of a Circle: Practice

Group and move the constant: $(x^2 - 4x) + (y^2 - 6y) = 12$. Half of $-4$ is $-2$ (square $4$), and half of $-6$ is $-3$ (square $9$); add both to each side.

$(x - 2)^2 + (y - 3)^2 = 12 + 4 + 9 = 25$

The center is $(2, 3)$ and the radius is $\sqrt{25} = 5$.

Complete the square: half of $4$ is $2$ (square $4$), half of $-2$ is $-1$ (square $1$).

$(x + 2)^2 + (y - 1)^2 = -5 + 4 + 1 = 0$

A sum of two squares equals $0$ only when each is $0$, forcing $x = -2$ and $y = 1$, the single point $(-2, 1)$.

Complete the square: half of $-2$ is $-1$ (square $1$), half of $6$ is $3$ (square $9$).

$(x - 1)^2 + (y + 3)^2 = -15 + 1 + 9 = -5$

A sum of two squares can never be negative, so no real point satisfies the equation and there is no graph.

The $x^2$ and $y^2$ coefficients are both $2$, so divide the whole equation by $2$ first: $x^2 + y^2 - 4x + 6y - 3 = 0$. Then complete the square.

$(x - 2)^2 + (y + 3)^2 = 3 + 4 + 9 = 16$

The center is $(2, -3)$.

First find $r^2$ as the squared distance from $(3, -2)$ to $(6, 2)$: $r^2 = 3^2 + 4^2 = 25$. Write the standard form and expand.

$(x - 3)^2 + (y + 2)^2 = 25 \Rightarrow x^2 - 6x + 9 + y^2 + 4y + 4 = 25$

Moving everything to one side gives $x^2 + y^2 - 6x + 4y - 12 = 0$.

The center sits halfway along a diameter, so average the two endpoints' coordinates.

$\left(\frac{1 + 7}{2}, \frac{2 + 10}{2}\right) = (4, 6)$

The center is $(4, 6)$.

There is no $y$ linear term, so $k = 0$. Complete the square on $x^2 + 6x$: half of $6$ is $3$ (square $9$).

$(x + 3)^2 + y^2 = 0 + 9 = 9$

The center is $(-3, 0)$ and the radius is $\sqrt{9} = 3$.

A circle needs an $x^2$ and a $y^2$ term with equal coefficients and the same sign. Check each option.

$x^2 - y^2 = 1 \text{ (opposite signs)}, \quad y = x^2 + 1 \text{ (no } y^2), \quad x^2 + 2y^2 = 4 \text{ (unequal coefficients)}$

Only $x^2 + y^2 - 4x + 2y + 1 = 0$ has matching $x^2$ and $y^2$ terms, so it is the circle.

Substitute $(a, 0)$ into the equation and solve for the positive $a$.

$a^2 + 0^2 = 169 \Rightarrow a^2 = 169 \Rightarrow a = 13$

The positive value is $a = 13$.

Test each point in the equation. A point on the circle makes the left side equal $25$.

$(0 + 4)^2 + (0 - 3)^2 = 16 + 9 = 25 \checkmark$

The origin works. The point $(-4, 3)$ is the center (left side $0$), and the others give $64$ and $16$.

Completing the square turns $x^2 + Dx$ into $\left(x + \tfrac{D}{2}\right)^2$ and $y^2 + Ey$ into $\left(y + \tfrac{E}{2}\right)^2$.

$\left(x + \tfrac{D}{2}\right)^2 + \left(y + \tfrac{E}{2}\right)^2 = \ldots \Rightarrow \text{center } \left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$

Each plus sign inside flips to give a negative center coordinate.

The center is equidistant from all three points. It is halfway between $(0, 0)$ and $(6, 0)$, so its $x$ is $3$, and halfway between $(0, 0)$ and $(0, 8)$, so its $y$ is $4$.

$\text{center} = (3, 4), \qquad r = \sqrt{3^2 + 4^2} = 5$

The center is $(3, 4)$, exactly $5$ from each of the three points.