Circles: Practice

Group the variables and move the constant across: $(x^2 + 6x) + (y^2 - 10y) = -18$.

Half of $6$ is $3$ (square $9$), and half of $-10$ is $-5$ (square $25$). Add both constants to both sides.

$(x + 3)^2 + (y - 5)^2 = -18 + 9 + 25 = 16$

The form subtracts the center coordinates, so $x + 3$ gives $h = -3$ and $y - 5$ gives $k = 5$. The radius is $\sqrt{16} = 4$, not the $16$ on the right.

The squared terms share the coefficient $3$, and every center-radius formula assumes that coefficient is $1$, so divide the whole equation by $3$ first.

$x^2 + y^2 - 4x + 6y - 5 = 0$

Now complete both squares. Half of $-4$ is $-2$ (square $4$), and half of $6$ is $3$ (square $9$).

$(x - 2)^2 + (y + 3)^2 = 5 + 4 + 9 = 18$

The center is $(2, -3)$ and the radius is $\sqrt{18} = 3\sqrt{2}$. Skipping the division at the start corrupts every number after it.

The center is the midpoint of the diameter, so average the coordinates: $\left(\frac{-2 + 4}{2}, \frac{3 + 11}{2}\right) = (1, 7)$.

The diameter is the distance between the endpoints, with horizontal gap $6$ and vertical gap $8$, so the radius is half of it.

$d = \sqrt{6^2 + 8^2} = 10 \;\Longrightarrow\; r = 5 \;\Longrightarrow\; r^2 = 25$

The circle is $(x - 1)^2 + (y - 7)^2 = 25$. Using the full distance $10$ as the radius gives $100$ on the right, a circle four times too large in area.

The radius is the distance from the center to the given point, and the equation only ever needs $r^2$, so stop before the square root.

$r^2 = (2 - (-1))^2 + (8 - 4)^2 = 3^2 + 4^2 = 25$

Substitute the center $(-1, 4)$ into the center-radius form, remembering that the form subtracts: $x - (-1) = x + 1$.

$(x + 1)^2 + (y - 4)^2 = 25$

The point $(2, 8)$ is on the circle, not its center, so it must not appear inside the parentheses.

Tangent to the $x$-axis means the circle touches that line at exactly one point, so the radius must equal the perpendicular distance from the center down to the axis.

That distance is the size of the center's $y$-coordinate.

$r = |-5| = 5$

The radius is $5$. The value $\sqrt{34}$ is the distance from the center to the origin, which is a different measurement entirely.

Every point on the $y$-axis has $x = 0$, so substitute $x = 0$ and solve for $y$.

$(0 - 3)^2 + (y - 4)^2 = 25 \;\Longrightarrow\; (y - 4)^2 = 16 \;\Longrightarrow\; y - 4 = \pm 4$

So $y = 8$ or $y = 0$, giving the points $(0, 0)$ and $(0, 8)$. The trap is to write $y - 4 = \pm 5$ using the radius, but the $9$ contributed by the $x$-term has already eaten part of the $25$.

The center of a general form sits at $\left(-\frac{D}{2}, -\frac{E}{2}\right)$, and here $D = -6$ and $E = 4$.

$\left(-\frac{-6}{2},\; -\frac{4}{2}\right) = (3, -2)$

A circle with that center is written $(x - 3)^2 + (y + 2)^2 = r^2$, with any positive $r^2$. Only $(x - 3)^2 + (y + 2)^2 = 1$ carries both the $x - 3$ and the $y + 2$.

Completing the square on $x^2 + y^2 + Dx + Ey + F = 0$ always leaves the center at $\left(-\frac{D}{2}, -\frac{E}{2}\right)$. Here $D = -7$ and $E = 1$, so halve each and flip the sign.

$\left(-\frac{-7}{2},\; -\frac{1}{2}\right) = \left(\frac{7}{2}, -\frac{1}{2}\right)$

Odd coefficients are no obstacle: the halves simply come out as fractions rather than whole numbers.

First find the radius by completing both squares. Half of $-4$ is $-2$ (square $4$), and half of $-6$ is $-3$ (square $9$).

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

So $r = \sqrt{25} = 5$, and the enclosed area is $\pi r^2$, which is the right side of the equation multiplied by $\pi$.

$A = \pi (5)^2 = 25\pi$

Treating the right side $25$ as the radius itself would give $625\pi$, so keep the square and the square root straight.

Complete the square. Half of $10$ is $5$ (square $25$), and half of $-24$ is $-12$ (square $144$).

$(x + 5)^2 + (y - 12)^2 = 0 + 25 + 144 = 169$

The radius is $\sqrt{169} = 13$, so the diameter is twice that.

$d = 2(13) = 26$

The constant term is $0$, so the circle passes through the origin, and the distance from the center $(-5, 12)$ to the origin is indeed $13$.

Complete the square. Half of $4$ is $2$ (square $4$), and half of $-6$ is $-3$ (square $9$), so the equation becomes $(x + 2)^2 + (y - 3)^2 = 13 - k$.

A genuine circle needs the right side to be strictly positive, since that right side is $r^2$.

$13 - k > 0 \;\Longrightarrow\; k < 13$

At $k = 13$ the graph is the single point $(-2, 3)$, not a circle, so the boundary value is excluded. Above $13$ there is no graph at all.

Tangent to the $y$-axis means the center sits a horizontal distance $4$ from it, and tangent to the $x$-axis means it sits a vertical distance $4$ from that. In the second quadrant $x$ is negative and $y$ is positive, so the center is $(-4, 4)$.

$(x - (-4))^2 + (y - 4)^2 = 4^2$

Simplify the double negative and square the radius.

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