When you substitute a line into a curve and the quadratic has a repeated root, the line touches the curve at exactly one point, so it is a tangent. The discriminant b² − 4ac is zero in that case.
This lesson follows eliminating a variable from a nonlinear pair and prepares the parameter questions in checking parameter values for a requested number of intersections. The discriminant itself is covered in quadratic structure and discriminants.
What does each outcome look like?
After substitution you have a quadratic in x. Its discriminant sorts the picture into three cases.
| Discriminant | Roots | Line and curve |
|---|---|---|
| b² − 4ac > 0 | two different | meet at two points |
| b² − 4ac = 0 | one repeated | touch at one point (tangent) |
| b² − 4ac < 0 | none real | do not meet |
A repeated root is the boundary between crossing and missing. Slide a line steadily across a circle and, for one moment, it touches.
How do you use it?
- Substitute and collect into ax² + bx + c = 0.
- Find the discriminant, or factorise and look for a perfect square such as (x + 2)².
- Read the geometry from the table above.
- For a tangent, find the point: solve the repeated root, then use the line to find y.
Worked example
Show that y = x + 4 is a tangent to x² + y² = 8, and find the point of contact.
Substitute: x² + (x + 4)² = 8.
Expand and collect: x² + x² + 8x + 16 = 8, so 2x² + 8x + 8 = 0, then x² + 4x + 4 = 0.
Discriminant: b² − 4ac = 16 − 16 = 0. The factorised form agrees: (x + 2)² = 0, so x = −2 twice.
Point of contact: y = −2 + 4 = 2, so the point is (−2, 2).
Check: (−2)² + 2² = 4 + 4 = 8, so the point is on the circle. The radius to (−2, 2) has gradient 2/(−2) = −1, the line has gradient 1, and −1 × 1 = −1, so they are perpendicular, as a tangent and radius should be.
The mistake to watch for
Students meet (x + 2)² = 0 and feel the answer is unfinished, so they invent a second root.
Mistaken working: (x + 2)² = 0, so x = −2 or x = 2.
The student used the ± habit from x² = 4, which does not apply to a square equal to zero.
If you test x = 2, then y = 6 and 4 + 36 = 40, not 8, so the point is not on the circle.
The correct reading is that (x + 2)² = 0 means x + 2 = 0 and nothing else. The single answer is not a missing answer. It is the algebra telling you the line touches the circle once.
Check yourself
Work these on paper, then open each answer.
1. Show that y = 4x − 4 touches the curve y = x², and find the point of contact.
Show answer
Set x² = 4x − 4, so x² − 4x + 4 = 0 and (x − 2)² = 0. The repeated root is x = 2, so the line is a tangent. Then y = 4(2) − 4 = 4.
Point of contact: (2, 4). Check: 2² = 4.
2. Does y = x + 6 touch x² + y² = 18? If so, where?
Show answer
x² + (x + 6)² = 18 gives 2x² + 12x + 36 = 18, so 2x² + 12x + 18 = 0 and x² + 6x + 9 = 0. That is (x + 3)² = 0, so x = −3 is repeated.
Yes, it is a tangent, at y = −3 + 6 = 3, so (−3, 3). Check: 9 + 9 = 18.
3. How many points do y = x + 5 and x² + y² = 18 share?
Show answer
x² + (x + 5)² = 18 gives 2x² + 10x + 25 = 18, so 2x² + 10x + 7 = 0. The discriminant is 100 − 4(2)(7) = 100 − 56 = 44, which is positive.
Two points. The line crosses the circle twice because the discriminant is greater than zero.
Where this leads next
Next, see how to turn words into equations in forming a system from two constraints, and then return to tangents with an unknown constant in checking parameter values. The line and circle intersection explorer shows a tangent as the moment two intersection points merge into one, and the quadratic structure explorer shows a double root as a graph touching the axis.
Some students can compute the discriminant but cannot say what it means in the picture. Our teachers work on that link in online one-to-one Additional Mathematics tuition.