Circles: Practice

Tangency means the line touches the circle exactly once, so the quadratic you get by substituting must have a repeated root, which happens exactly when its discriminant is zero.

$x^2 + (x + c)^2 = 9 \;\Longrightarrow\; 2x^2 + 2cx + (c^2 - 9) = 0$

Set the discriminant to zero and solve for $c$.

$(2c)^2 - 4(2)(c^2 - 9) = -4c^2 + 72 = 0 \;\Longrightarrow\; c^2 = 18$

So $c = \pm\sqrt{18} = \pm 3\sqrt{2}$. Two lines qualify, one above the circle and one below, which is what any picture of a circle demands.

Substitute the line into the circle and collect a quadratic in $x$.

$x^2 + (-x + 6)^2 = 16 \;\Longrightarrow\; 2x^2 - 12x + 20 = 0 \;\Longrightarrow\; x^2 - 6x + 10 = 0$

The discriminant decides the count.

$(-6)^2 - 4(1)(10) = 36 - 40 = -4 < 0$

A negative discriminant means no real roots, so the line misses the circle completely. The circle has radius $4$, and this line passes too far from the origin to reach it.

Use the general form $x^2 + y^2 + Dx + Ey + F = 0$, which is linear in $D$, $E$, and $F$, and substitute each point. The origin gives $F = 0$ immediately.

With $F = 0$, the point $(4, 0)$ gives $16 + 4D = 0$, so $D = -4$, and the point $(0, -6)$ gives $36 - 6E = 0$, so $E = 6$.

$x^2 + y^2 - 4x + 6y = 0 \;\Longrightarrow\; (x - 2)^2 + (y + 3)^2 = 4 + 9 = 13$

Check all three points against the center $(2, -3)$: each sits a squared distance of $13$ away, since $4 + 9 = 13$ in every case.

A common point satisfies both equations, so it also satisfies their difference. Write the first as $x^2 + y^2 - 25 = 0$ and subtract the second, which cancels the $x^2$ and $y^2$ terms and leaves a linear equation.

$(x^2 + y^2 - 25) - (x^2 + y^2 - 6x - 8y + 9) = 6x + 8y - 34 = 0$

Divide by $2$ to simplify.

$3x + 4y = 17$

Every shared point lies on this line, so it is the line through both intersection points. The squared terms always cancel this way, which is why two distinct circles meet in at most two points: their shared points all sit on one line, and a line meets a circle at most twice.

Substitute $y = 3$ into the circle to find where the line crosses it.

$x^2 + 3^2 = 25 \;\Longrightarrow\; x^2 = 16 \;\Longrightarrow\; x = \pm 4$

The endpoints are $(-4, 3)$ and $(4, 3)$. They share a $y$-coordinate, so the chord is horizontal and its length is the gap between the $x$-values.

$4 - (-4) = 8$

The chord is $8$ units long, shorter than the diameter $10$, as any chord other than a diameter must be.

The circle has center $(6, 8)$ and radius $\sqrt{4} = 2$. First measure how far the center is from the origin.

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

Every point of the circle is exactly $2$ from that center, so its distance from the origin can range from $10 - 2$ to $10 + 2$. The far point is the one on the ray from the origin straight through the center, extended past it.

$10 + 2 = 12$

The greatest distance is $12$, and (for contrast) the least is $8$.

Substitute $y = mx$ into the circle and collect a quadratic in $x$.

$(x - 5)^2 + m^2x^2 = 9 \;\Longrightarrow\; (1 + m^2)x^2 - 10x + 16 = 0$

Tangency means a repeated root, so set the discriminant to zero.

$(-10)^2 - 4(1 + m^2)(16) = 100 - 64 - 64m^2 = 36 - 64m^2 = 0$

Solving gives $m^2 = \frac{36}{64} = \frac{9}{16}$, so $m = \pm\frac{3}{4}$. The value $\frac{9}{16}$ is $m^2$, not $m$, which is the trap on this item.

Read the centers and radii: $(0, 0)$ with $r_1 = 1$, and $(5, 0)$ with $r_2 = \sqrt{16} = 4$. The centers are $5$ apart.

Compare that separation with the sum of the radii.

$r_1 + r_2 = 1 + 4 = 5 = \text{distance between the centers}$

The circles reach each other exactly, touching at one point and no more: they are externally tangent, meeting at $(1, 0)$.

Watch both ends of that comparison. A larger separation and the circles would miss. A slightly smaller one and they would cross twice, but only until the separation drops below the difference of the radii, $4 - 1 = 3$. Below that the small circle is swallowed by the large one and they share no points again, so closer does not always mean more contact.

The squared terms carry a common coefficient of $4$, so divide the entire equation by $4$ before completing the square.

$x^2 + y^2 - x + 4y - \frac{19}{4} = 0$

Half of $-1$ is $-\frac{1}{2}$ (square $\frac{1}{4}$), and half of $4$ is $2$ (square $4$). Add both to each side.

$\left(x - \frac{1}{2}\right)^2 + (y + 2)^2 = \frac{19}{4} + \frac{1}{4} + 4 = 5 + 4 = 9$

The radius is $\sqrt{9} = 3$, and the center is $\left(\frac{1}{2}, -2\right)$.

Passing through the point is not enough, since every choice does that. A tangent meets the circle only there, so the quadratic you get by substituting must have a repeated root.

Test $3x + 4y = 25$ by writing $y = \frac{25 - 3x}{4}$, substituting, and clearing the denominator.

$16x^2 + (25 - 3x)^2 = 400 \;\Longrightarrow\; 25x^2 - 150x + 225 = 0 \;\Longrightarrow\; (x - 3)^2 = 0$

The repeated root $x = 3$ is the double-root signature of tangency, so this line touches at $(3, 4)$ and nowhere else.

The others all cut through. The line $y = \frac{4}{3}x$ runs along the radius and straight through the center, so it is a diameter. Substituting $4x + 3y = 24$ or $x + y = 7$ gives a positive discriminant in each case, so each crosses the circle twice. The tangent is the one perpendicular to the radius at $(3, 4)$: that radius has slope $\frac{4}{3}$, so the tangent's slope is $-\frac{3}{4}$, which is exactly the slope of $3x + 4y = 25$.

Substitute the line into the circle and collect a quadratic in $x$.

$x^2 + (2x + k)^2 = 5 \;\Longrightarrow\; 5x^2 + 4kx + (k^2 - 5) = 0$

No intersection means no real roots, so demand a negative discriminant.

$(4k)^2 - 4(5)(k^2 - 5) = -4k^2 + 100 < 0 \;\Longrightarrow\; k^2 > 25$

So $|k| > 5$, that is $k < -5$ or $k > 5$. At $k = \pm 5$ the discriminant is zero and the line is tangent, which is a touch, not a miss, so those two values are excluded.

Expand both squares and move the $4$ across, so the constant becomes $9 + 1 - 4$.

$x^2 - 6x + 9 + y^2 - 2y + 1 = 4 \;\Longrightarrow\; x^2 + y^2 - 6x - 2y + 6 = 0$

The other choices all differ. A constant of $10$ gives $r^2 = 0$, a single point; a constant of $-4$ gives $r^2 = 14$; and flipping the signs of the linear terms moves the center to $(-3, -1)$, a different circle of the same size.