To find the values of k for which a line meets a curve a given number of times, substitute, find the discriminant of the resulting quadratic, and set it greater than, equal to or less than zero. Greater than zero gives two intersections, equal to zero gives one (a tangent) and less than zero gives none.
This lesson follows interpreting a repeated solution geometrically and closes the lesson sequence of simultaneous linear and nonlinear models. The discriminant is revised in quadratic structure and discriminants.
What changes when the line contains k?
Nothing in the method changes, but the coefficients now contain k, so the discriminant becomes an expression in k rather than a number. The inequality is then solved like any quadratic inequality in k.
Because the quadratic’s coefficients depend on k, be careful to expand the bracket in full before finding b² − 4ac. Most errors here come from the middle term.
How do you do it, step by step?
- Substitute the line into the curve.
- Expand and collect to ax² + bx + c = 0, with a, b and c possibly containing k.
- Write b² − 4ac in terms of k and simplify.
- Apply the condition: > 0 for two points, = 0 for one, < 0 for none.
- Solve for k and state the interval or the values.
- Test one value of k to confirm the result.
Worked example
The line y = x + k and the circle x² + y² = 8. Find the values of k for which the line meets the circle at two distinct points, at one point, and at no points.
Substitute: x² + (x + k)² = 8.
Expand and collect: x² + x² + 2kx + k² = 8, so 2x² + 2kx + (k² − 8) = 0.
Discriminant: a = 2, b = 2k, c = k² − 8.
b² − 4ac = 4k² − 8(k² − 8) = 4k² − 8k² + 64 = 64 − 4k² = 4(16 − k²).
Conditions:
| Case | Condition | Values of k |
|---|---|---|
| two points | 16 − k² > 0 | −4 < k < 4 |
| one point | 16 − k² = 0 | k = 4 or k = −4 |
| no points | 16 − k² < 0 | k < −4 or k > 4 |
Check with distance: the line x − y + k = 0 is |k|/√2 from the origin. The radius is √8 = 2√2. Touching needs |k|/√2 = 2√2, so |k| = 4, which matches. For k = 2 the distance is √2, less than 2√2, so two points. For k = 5 the distance is about 3.54, more than 2.83, so no points.
The case k = 4 is the tangent y = x + 4 met in the previous lesson, where the contact point was (−2, 2).
The mistake to watch for
A frequent error is to solve the inequality as if it were linear.
Mistaken working: 16 − k² > 0, so k² < 16, so k < 4.
The student took the positive square root only and dropped the lower boundary.
Test it: k = −10 satisfies k < 4, but the line y = x − 10 is nowhere near the circle, since 10/√2 is about 7.07, far more than 2.83. The correction is to remember that k² < 16 means the interval between −4 and 4. For no intersections, k² > 16 gives two separate pieces, k < −4 or k > 4, and not one.
Check yourself
Work these on paper, then open each answer.
1. Find the values of c for which y = 2x + c meets y = x² at two distinct points.
Show answer
Set x² = 2x + c, so x² − 2x − c = 0. The discriminant is 4 + 4c. For two points, 4 + 4c > 0, so c > −1.
Check at c = −1: x² − 2x + 1 = 0 gives (x − 1)² = 0, a tangent at (1, 1).
2. The line y = x + k touches the circle x² + y² = 2. Find k.
Show answer
x² + (x + k)² = 2 gives 2x² + 2kx + k² − 2 = 0. The discriminant is 4k² − 8(k² − 2) = 16 − 4k². Setting it to zero gives k² = 4.
k = 2 or k = −2. Check by distance: |k|/√2 = √2 gives |k| = 2.
3. Find the values of c for which y = 2x + c does not meet x² + y² = 5.
Show answer
x² + (2x + c)² = 5 gives 5x² + 4cx + c² − 5 = 0. The discriminant is 16c² − 20(c² − 5) = 100 − 4c². For no intersection it must be negative, so c² > 25.
c < −5 or c > 5. Check c = 6: the distance from the origin is 6/√5 ≈ 2.68, more than √5 ≈ 2.24.
Where this leads next
After this lesson, try the mixed practice set for questions that combine all five lessons. The line and circle intersection explorer lets you see the tangent positions as k changes, and the non-calculator working trainer supports exact simplification of expressions such as 4(16 − k²).
If inequality steps with an unknown constant keep costing you marks, our teachers can rebuild them with you in online one-to-one Additional Mathematics tuition.