In the xy-plane, the graph of the equation above is a circle. Point P is on the circle and has...

1. TRANSLATE the circle equation and given information

• Given information:
  • Circle equation: \((x - 6)^2 + (y + 5)^2 = 16\)
  • Point P(10, -5) is on the circle
  • PQ is a diameter (so Q is the other endpoint)

• From standard form \((x - h)^2 + (y - k)^2 = r^2\):
  • Center: (6, -5)
  • Radius: \(\sqrt{16} = 4\)

2. INFER the key geometric relationship

• Since PQ is a diameter, it passes through the center
• The center (6, -5) is exactly halfway between P(10, -5) and Q
• This means we can use the midpoint formula to find Q

3. TRANSLATE the midpoint relationship into equations

• If Q has coordinates (x, y), then:

Center = Midpoint of P and Q

\((6, -5) = \left(\frac{10 + x}{2}, \frac{-5 + y}{2}\right)\)

4. SIMPLIFY to solve for Q's coordinates

• For the x-coordinate:

\(6 = \frac{10 + x}{2}\)

\(12 = 10 + x\)

\(x = 2\)

• For the y-coordinate:

\(-5 = \frac{-5 + y}{2}\)

\(-10 = -5 + y\)

\(y = -5\)

Answer: A. (2, -5)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skills: Students may correctly identify the center as (6, -5) but misunderstand what "PQ is a diameter" means geometrically. They might think Q is just any other point on the circle rather than the specific point that makes PQ pass through the center.

Without understanding that the center is the midpoint of the diameter, students get stuck and resort to guessing, often selecting Choice C (6, -5) because they recognize it as the center.

Second Most Common Error:

Poor SIMPLIFY execution: Students understand the midpoint concept but make algebraic errors when solving the equations. A common mistake is incorrectly solving \(6 = \frac{10 + x}{2}\), perhaps getting \(x = -2\) instead of \(x = 2\), or making sign errors with the y-coordinate calculation.

This computational error might lead them to select Choice D (6, -9) if they get the y-coordinate wrong.

The Bottom Line:

This problem tests whether students can connect the geometric concept of a diameter with the algebraic tool of the midpoint formula. Success requires both conceptual understanding and careful algebraic manipulation.