13.2 Equations of lines and circles in the coordinate plane

Equations of Lines

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Equations of lines from given information

Different forms of a line's equation highlight different information. Choosing the right form depends on what you're given and what you need to find.

Slope-intercept form: $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. This is the most common form and the easiest to graph from, since you can plot the y-intercept and then use the slope to find additional points.

Point-slope form: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is any known point on the line. Use this when you're given a slope and a point (or can calculate the slope) but don't know the y-intercept yet. It's also the go-to form for writing equations of parallel and perpendicular lines through a specific point.

Two-point form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$, where $(x_1, y_1)$ and $(x_2, y_2)$ are two known points. This is really just point-slope form with the slope formula plugged in. You calculate the slope from the two points and then use either point to write the equation.

Parallel and perpendicular slopes come up constantly in coordinate geometry proofs and problems:

• Parallel lines have equal slopes: $m_1 = m_2$
• Perpendicular lines have slopes that are negative reciprocals: $m_1 \cdot m_2 = -1$
  • For example, if one line has slope $2$, a line perpendicular to it has slope $-\frac{1}{2}$, because $2 \cdot (-\frac{1}{2}) = -1$.
  • Watch out for horizontal and vertical lines. A horizontal line (slope $0$) is perpendicular to a vertical line (undefined slope), but the negative reciprocal rule doesn't apply in that case since you can't take the reciprocal of $0$.

Standard form: $Ax + By = C$, where $A$, $B$, and $C$ are integers and $A \geq 0$. You'll sometimes be asked to give your final answer in this form. To convert from slope-intercept or point-slope, clear any fractions by multiplying through, then rearrange so both variable terms are on the left and the constant is on the right.

Equations of Circles

Center and radius of circles

A circle is the set of all points equidistant from a fixed center point. That distance is the radius, and the equation captures both pieces of information.

Standard form: $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

Pay close attention to the signs. The formula uses subtraction, so a center of $(3, -2)$ produces $(x - 3)^2 + (y - (-2))^2$, which simplifies to $(x - 3)^2 + (y + 2)^2$. The plus sign in front of the $2$ means the actual y-coordinate of the center is $-2$. A common mistake is reading $(y + 2)$ and thinking $k = 2$. Always flip the sign you see inside the parentheses.

• Example: $(x - 3)^2 + (y + 2)^2 = 16$ has center $(3, -2)$ and radius $4$ (since $\sqrt{16} = 4$).
• Example: $(x + 1)^2 + (y - 5)^2 = 7$ has center $(-1, 5)$ and radius $\sqrt{7}$. The radius doesn't have to be a whole number.

Circle equations: standard vs general form

General form: $x^2 + y^2 + Ax + By + C = 0$

This form isn't as immediately useful because you can't read off the center or radius. To extract that information, you need to convert to standard form by completing the square.

Converting general form to standard form:

1. Group the $x$ terms together and the $y$ terms together. Move the constant to the right side.
2. Complete the square for the $x$ group: take half the coefficient of $x$, square it, and add it to both sides.
3. Complete the square for the $y$ group the same way.
4. Factor each group into a squared binomial. The right side equals $r^2$.

Example: Convert $x^2 + y^2 - 2x + 6y + 1 = 0$ to standard form.

1. Group and move the constant: $(x^2 - 2x) + (y^2 + 6y) = -1$

2. Complete the square for $x$: half of $-2$ is $-1$, and $(-1)^2 = 1$. Add $1$ to both sides.

3. Complete the square for $y$: half of $6$ is $3$, and $3^2 = 9$. Add $9$ to both sides.

4. Factor: $(x - 1)^2 + (y + 3)^2 = -1 + 1 + 9 = 9$

So the circle has center $(1, -3)$ and radius $3$.

Quick check: After completing the square, $r^2$ must be positive. If you get $r^2 = 0$, the "circle" is just a single point. If you get $r^2 < 0$, the equation has no graph at all, meaning something went wrong or the equation doesn't represent a real circle.

Converting standard form to general form is simpler: expand the squared binomials, combine like terms, and set the equation equal to zero.

Tangent lines to circles

A tangent line touches a circle at exactly one point, called the point of tangency. The key geometric fact is that the radius drawn to the point of tangency is always perpendicular to the tangent line. This perpendicularity is what lets you find the tangent line's equation.

Steps to find the equation of a tangent line:

1. Find the slope of the radius by using the slope formula with the center and the point of tangency.
2. Take the negative reciprocal of that slope. That's the slope of the tangent line.
3. Use point-slope form with the point of tangency and the tangent slope to write the equation.

Example: Find the tangent line to a circle with center $(2, 3)$ at the point of tangency $(5, 6)$.

1. Slope of the radius: $\frac{6 - 3}{5 - 2} = \frac{3}{3} = 1$

2. Slope of the tangent line: $-1$ (negative reciprocal of $1$)

3. Using point-slope form: $y - 6 = -1(x - 5)$, which simplifies to $y = -x + 11$

Special cases to watch for: If the radius is horizontal (slope $0$), the tangent line is vertical, so its equation is $x = x_1$. If the radius is vertical (undefined slope), the tangent line is horizontal, so its equation is $y = y_1$. These cases come up when the point of tangency is directly above, below, or to the side of the center.