A circle in the xy-plane has its center at \((-4, -6)\). Line k is tangent to this circle at the...

1. TRANSLATE the problem information

• Given information:
  • Circle center at (-4, -6)
  • Line k is tangent to circle at point (-7, -7)
  • Need to find slope of line k

2. INFER the geometric relationship

• Key insight: A tangent line to a circle is always perpendicular to the radius at the point of tangency
• This means we need to find the radius slope first, then use the perpendicular relationship

3. SIMPLIFY to find the radius slope

• The radius connects center (-4, -6) to tangent point (-7, -7)
• Using slope formula: \(\mathrm{m = \frac{y_2 - y_1}{x_2 - x_1}}\)
• Radius slope = \(\mathrm{\frac{-7 - (-6)}{-7 - (-4)}}\) = \(\mathrm{\frac{-1}{-3}}\) = \(\mathrm{\frac{1}{3}}\)

4. INFER the tangent line slope

• Since perpendicular lines have slopes that are negative reciprocals:
• Tangent slope = \(\mathrm{-\frac{1}{\text{radius slope}}}\) = \(\mathrm{-\frac{1}{\frac{1}{3}}}\) = \(\mathrm{-3}\)

Answer: A. -3

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Not knowing that tangent lines are perpendicular to radii at the point of tangency

Without this key geometric property, students may try to use the two given points directly as if they're on the same line, or they may attempt other approaches that don't lead to a solution. This leads to confusion and guessing.

Second Most Common Error:

Weak INFER skill: Calculating the radius slope correctly \(\mathrm{\frac{1}{3}}\) but failing to recognize they need the negative reciprocal for the perpendicular tangent line

This may lead them to select Choice C \(\mathrm{\frac{1}{3}}\) - which is actually the slope of the radius, not the tangent line.

The Bottom Line:

This problem requires connecting coordinate geometry with circle properties. Students must recognize the perpendicular relationship and then execute the negative reciprocal calculation correctly.