The line y = 2x + 9 intersects the circle x^2 + y^2 = 16 at how many distinct points?

1. TRANSLATE the problem setup

• Given information:
  • Line: \(\mathrm{y = 2x + 9}\)
  • Circle: \(\mathrm{x^2 + y^2 = 16}\)
  • Need: Number of intersection points

• This means finding points \(\mathrm{(x,y)}\) that satisfy both equations simultaneously

2. INFER the solution approach

• To find intersections, substitute the line equation into the circle equation
• This eliminates y and gives us a quadratic equation in x
• The number of real solutions for x tells us the number of intersection points

3. TRANSLATE by substitution

• Substitute \(\mathrm{y = 2x + 9}\) into \(\mathrm{x^2 + y^2 = 16}\):
  \(\mathrm{x^2 + (2x + 9)^2 = 16}\)

4. SIMPLIFY the equation

• Expand \(\mathrm{(2x + 9)^2}\):
  \(\mathrm{(2x + 9)^2 = 4x^2 + 36x + 81}\)
• Substitute:
  \(\mathrm{x^2 + 4x^2 + 36x + 81 = 16}\)
• Combine like terms:
  \(\mathrm{5x^2 + 36x + 81 = 16}\)
• Rearrange:
  \(\mathrm{5x^2 + 36x + 65 = 0}\)

5. INFER using the discriminant

• For quadratic \(\mathrm{ax^2 + bx + c = 0}\), discriminant \(\mathrm{D = b^2 - 4ac}\)
• Here: \(\mathrm{a = 5, b = 36, c = 65}\)
• Calculate:
  \(\mathrm{D = 36^2 - 4(5)(65) = 1296 - 1300 = -4}\)

6. INFER the final answer

• Since \(\mathrm{D = -4 \lt 0}\), the quadratic has no real solutions
• No real x-values means no intersection points

Answer: (D) Zero

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when expanding \(\mathrm{(2x + 9)^2}\), most commonly getting \(\mathrm{4x^2 + 18x + 81}\) instead of \(\mathrm{4x^2 + 36x + 81}\) (forgetting the middle term is \(\mathrm{2·2x·9 = 36x}\)).

This leads to \(\mathrm{5x^2 + 18x + 65 = 0}\), giving discriminant \(\mathrm{D = 18^2 - 4(5)(65) = 324 - 1300 = -976}\). While still negative (correct conclusion), calculation errors here could easily flip signs and lead to confusion about whether intersections exist.

This may lead them to select Choice (A) (Exactly one) if they incorrectly calculate a discriminant of zero, or cause confusion and guessing.

Second Most Common Error:

Missing conceptual knowledge about discriminant interpretation: Students might correctly set up and solve the quadratic but not remember what negative discriminant means for intersections.

They might think "negative discriminant = one intersection" or get confused about the relationship between discriminant sign and number of solutions. This leads them to abandon systematic solution and guess randomly among the choices.

The Bottom Line:

This problem requires students to connect algebraic techniques (substitution, quadratic analysis) with geometric concepts (line-circle intersections). The discriminant is the key bridge between algebra and geometry here.