Circles: Additional Practice (Set 1)

Move every variable term to the left so the equation is in general form: $x^2 + y^2 - 6x - 8y = 0$.

Now complete both squares. Half of $-6$ is $-3$ (square $9$), and half of $-8$ is $-4$ (square $16$).

$(x - 3)^2 + (y - 4)^2 = 0 + 9 + 16 = 25$

The center is $(3, 4)$ and the radius is $\sqrt{25} = 5$. Since the constant term is $0$, this circle passes through the origin, and indeed the center is $5$ units from it.

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

$x^2 - 2x - 8 = 0 \;\Longrightarrow\; (x - 4)(x + 2) = 0$

The roots are $x = 4$ and $x = -2$, giving the points $(4, 0)$ and $(-2, 0)$.

As a check, completing the square gives center $(1, 4)$ and radius $5$, and the distance from $(1, 4)$ to $(4, 0)$ is $\sqrt{9 + 16} = 5$, so that point really is on the circle.

Substitute $y = 4$ into the circle and see how many values of $x$ survive.

$x^2 + (4 - 1)^2 = 9 \;\Longrightarrow\; x^2 + 9 = 9 \;\Longrightarrow\; x^2 = 0$

The equation $x^2 = 0$ has the repeated root $x = 0$, which is the double-root signature of tangency. The line touches the circle at the single point $(0, 4)$.

That matches the picture: the center is $(0, 1)$ with radius $3$, so the highest point of the circle is exactly $(0, 4)$.

Read off the centers and radii: $(1, 2)$ with $r_1 = 2$, and $(7, 10)$ with $r_2 = 6$. Measure the distance between the centers.

$d = \sqrt{(7 - 1)^2 + (10 - 2)^2} = \sqrt{36 + 64} = 10$

Compare that with the sum of the radii, $2 + 6 = 8$. The centers are $10$ apart, farther than the two circles can reach toward each other.

$d = 10 > 8 = r_1 + r_2$

The circles lie entirely outside one another and share no points at all.

The center is $(8, 6)$ and the radius is $\sqrt{25} = 5$. Measure how far the center sits from the origin.

$\sqrt{8^2 + 6^2} = \sqrt{100} = 10$

Since $10 > 5$, the origin lies outside the circle. Points of the circle sit between $10 - 5$ and $10 + 5$ from the origin, and the nearest one is on the segment joining the origin to the center.

$10 - 5 = 5$

The least distance is $5$. It is not $0$, because the origin is not on the circle: substituting gives $64 + 36 = 100$, not $25$.

Complete the square to see the circle plainly. Half of $-6$ is $-3$ (square $9$), and half of $-4$ is $-2$ (square $4$).

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

The center is $(3, 2)$ and the radius is $5$, so the circle occupies the horizontal band from $y = 2 - 5$ up to $y = 2 + 5$. Substituting $y = k$ leaves $(x - 3)^2 = 25 - (k - 2)^2$, which has two real solutions exactly when the right side is strictly positive.

$(k - 2)^2 < 25 \;\Longrightarrow\; -3 < k < 7$

At $k = -3$ and $k = 7$ the line is tangent, giving one point, not two, so those endpoints are excluded.

Substitute the three points into $x^2 + y^2 + Dx + Ey + F = 0$. From $(2, 0)$: $4 + 2D + F = 0$. From $(-2, 0)$: $4 - 2D + F = 0$. Subtracting these gives $4D = 0$, so $D = 0$, and then $F = -4$.

From $(0, 2)$: $4 + 2E + F = 0$, and with $F = -4$ this gives $E = 0$.

$x^2 + y^2 - 4 = 0 \;\Longrightarrow\; x^2 + y^2 = 4$

The circle is centered at the origin with radius $2$, and each of the three given points is exactly $2$ units from $(0, 0)$.

Every choice has $D = 2$ and $E = -6$, so completing the square always gives $(x + 1)^2 + (y - 3)^2 = 10 - F$, and only the constant changes.

A sum of two squares is zero only when both squares vanish, so the graph is a single point exactly when the right side is $0$.

$10 - F = 0 \;\Longrightarrow\; F = 10$

With $F = 10$ the lone solution is $(-1, 3)$. The constants $6$ and $0$ leave positive right sides (circles), and $14$ leaves a negative one (no graph at all).

Substitute $y = 6$ into the circle to locate the endpoints.

$x^2 + 36 = 100 \;\Longrightarrow\; x^2 = 64 \;\Longrightarrow\; x = \pm 8$

The endpoints are $(-8, 6)$ and $(8, 6)$, which share a $y$-coordinate, so the chord is horizontal and its length is the gap between the $x$-values.

$8 - (-8) = 16$

The chord is $16$ long. Answering $8$ would report only the half-chord, which is the leg of the right triangle with hypotenuse $10$ and other leg $6$.

The squared terms share the coefficient $9$, so divide the whole equation by $9$ before anything else.

$x^2 + y^2 - \frac{2}{3}x + 2y - \frac{26}{9} = 0$

Complete both squares. Half of $-\frac{2}{3}$ is $-\frac{1}{3}$ (square $\frac{1}{9}$), and half of $2$ is $1$ (square $1$).

$\left(x - \frac{1}{3}\right)^2 + (y + 1)^2 = \frac{26}{9} + \frac{1}{9} + 1 = 3 + 1 = 4$

The center is $\left(\frac{1}{3}, -1\right)$ and the radius is $\sqrt{4} = 2$. Fractions in the center are perfectly normal; what is not allowed is skipping the division by $9$.

The center of a general form sits at $\left(-\frac{D}{2}, -\frac{E}{2}\right)$, so with $D = -8$ and $E = 6$ the center is $(4, -3)$.

Build the line through $(4, -3)$ with slope $2$.

$y - (-3) = 2(x - 4) \;\Longrightarrow\; y = 2x - 11$

Check it: at $x = 4$ the line gives $y = 8 - 11 = -3$, which is the center. (For the record, the circle is real: $64 + 36 + 44 = 144 > 0$, so its radius is $6$.)

First put the circle in center-radius form: completing the square gives $(x - 2)^2 + (y - 1)^2 = 8 + 4 + 1 = 13$, so the center is $(2, 1)$ and the radius is $\sqrt{13}$.

Now test whether the line passes through that center. At $x = 2$ the line gives $y = 2 - 1 = 1$, which is exactly the center, so the chord is a diameter and its length is $2r$.

$2\sqrt{13}$

The substitution route agrees. Replacing $y$ gives $2x^2 - 8x - 5 = 0$, whose roots differ by $\sqrt{26}$; since the slope is $1$, the vertical gap matches the horizontal one, so the distance is $\sqrt{26 + 26} = \sqrt{52} = 2\sqrt{13}$.