Circles Pattern - Tangent Lines

This pattern asks you to find the slope of a tangent line to a circle, using the key property that a tangent line is perpendicular to the radius at the point of tangency.

The Perpendicular Radius Property

If a line is tangent to a circle at point $T$, and the circle has center $C$, then the radius $\overline{CT}$ is perpendicular to the tangent line at $T$. Perpendicular lines have slopes that are negative reciprocals of each other:

$m_{\text{tangent}} = -\dfrac{1}{m_{\text{radius}}}$

Finding the Tangent Slope

Step 1: Find the slope of the radius from the center to the tangency point.
Step 2: Take the negative reciprocal.

A circle has center $(2, -1)$. Line $t$ is tangent at $(5, 5)$. What is the slope of $t$?

A) $-2$
B) $-\dfrac{1}{2}$
C) $\dfrac{1}{2}$
D) $2$

Radius slope $= \dfrac{5 - (-1)}{5 - 2} = \dfrac{6}{3} = 2$. Tangent slope $= -\dfrac{1}{2}$. The answer is B.

A circle has center $(2, 5)$. Line $m$ is tangent at $(6, 3)$. What is the slope of $m$?

A) $-2$
B) $-\dfrac{1}{2}$
C) $\dfrac{1}{2}$
D) $2$

Radius slope $= \dfrac{3 - 5}{6 - 2} = \dfrac{-2}{4} = -\dfrac{1}{2}$. Tangent slope $= -\dfrac{1}{-1/2} = 2$. The answer is D.

A circle has center $(5, -2)$. Line $m$ is tangent at $(3, 2)$. What is the slope of $m$?

A) $-2$
B) $-\dfrac{1}{2}$
C) $\dfrac{1}{2}$
D) $2$

Radius slope $= \dfrac{2 - (-2)}{3 - 5} = \dfrac{4}{-2} = -2$. Tangent slope $= -\dfrac{1}{-2} = \dfrac{1}{2}$. The answer is C.

A circle has center $(3, -5)$. Line $m$ is tangent at $(1, 1)$. What is the slope?

A) $-3$
B) $-\dfrac{1}{3}$
C) $3$
D) $\dfrac{1}{3}$

Radius slope $= \dfrac{1 - (-5)}{1 - 3} = \dfrac{6}{-2} = -3$. Tangent slope $= \dfrac{1}{3}$. The answer is D.

Identifying a Point on the Tangent Line

Some problems give the center, the tangency point, and ask which of four points lies on the tangent line. First find the tangent slope, then check which point gives the same slope when paired with the tangency point.

A circle has center $(2, -3)$. Line $k$ is tangent at $(5, 1)$. Which point also lies on $k$?

A) $(8, -3)$
B) $(9, -2)$
C) $(9, 4)$
D) $(6, -6)$

Radius slope $= \dfrac{1 - (-3)}{5 - 2} = \dfrac{4}{3}$. Tangent slope $= -\dfrac{3}{4}$.

Check each option using slope from $(5, 1)$: - A) $\dfrac{-3 - 1}{8 - 5} = \dfrac{-4}{3}$ — close but positive/negative differs - B) $\dfrac{-2 - 1}{9 - 5} = \dfrac{-3}{4} = -\dfrac{3}{4}$ — matches!

The answer is B.

A circle has center $(2, 5)$. Line $k$ is tangent at $(-1, -1)$. Which point also lies on $k$?

A) $(1, 3)$
B) $(5, -4)$
C) $(3, 1)$
D) $(2, -7)$

Radius slope $= \dfrac{-1 - 5}{-1 - 2} = \dfrac{-6}{-3} = 2$. Tangent slope $= -\dfrac{1}{2}$.

Check from $(-1, -1)$: - B) $\dfrac{-4 - (-1)}{5 - (-1)} = \dfrac{-3}{6} = -\dfrac{1}{2}$ — matches!

The answer is B.

Tip: You don't need to check all four options. Once you find a match, move on. But if you want to verify, use the tangent slope to write the full line equation: $y - y_1 = m(x - x_1)$, then test each point.

What to Do on Test Day

• The one rule you need: A tangent line is perpendicular to the radius at the point of tangency. Perpendicular slopes are negative reciprocals: $m_1 \cdot m_2 = -1$.
• Two-step process: 1. Find the slope of the radius: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ (from center to tangent point). 2. Take the negative reciprocal: $m_{\text{tangent}} = -\dfrac{1}{m}$.
• Checking a point on the tangent line: Compute the slope from the tangent point to the candidate point. If it matches the tangent slope, the point is on the line.
• Common sign errors: Be careful with negative coordinates. $\dfrac{-6}{-3} = 2$, not $-2$. Write out the subtraction step by step.
• Horizontal and vertical tangents: If the radius is vertical (undefined slope), the tangent is horizontal (slope 0), and vice versa.