A line is a tangent to a circle when it meets the circle at exactly one point. To find a parameter such as k, substitute the line into the circle, collect a quadratic, and set its discriminant equal to zero.
This lesson joins finding intersections of a line and circle with the discriminant work in quadratic structure and discriminants. It belongs to circle coordinate methods.
What does each discriminant value mean?
After substitution you have a quadratic ax² + bx + c = 0 whose coefficients may contain k. The discriminant is b² − 4ac.
| Discriminant | Number of intersections | The line is |
|---|---|---|
| b² − 4ac > 0 | 2 | a secant (cuts the circle twice) |
| b² − 4ac = 0 | 1 | a tangent |
| b² − 4ac < 0 | 0 | outside the circle |
So the tangency condition turns a geometry question into an equation in k.
How do you do it, step by step?
- Substitute the line into the circle, keeping k as a letter.
- Expand with care, including the k-terms in the middle.
- Collect into ax² + bx + c = 0, where b and c may contain k.
- Set b² − 4ac = 0 and solve the equation for k, often a quadratic in k.
- Give every solution, then check one value by finding the point of contact or a distance.
Worked example
Find the values of k for which the line y = x + k is a tangent to the circle x² + y² = 8.
Step 1, substitute: x² + (x + k)² = 8.
Step 2, expand and collect: x² + x² + 2kx + k² − 8 = 0, so 2x² + 2kx + (k² − 8) = 0.
Step 3, discriminant: a = 2, b = 2k, c = k² − 8.
b² − 4ac = 4k² − 8(k² − 8) = 4k² − 8k² + 64 = 64 − 4k².
Step 4, set to zero: 64 − 4k² = 0, so k² = 16 and k = 4 or k = −4.
Check: for k = 4 the quadratic is 2x² + 8x + 8 = 0, which gives (x + 2)² = 0, so x = −2 and y = 2. The point (−2, 2) is on the circle since 4 + 4 = 8. ✓ The distance method agrees: |k| ÷ √2 = 2√2 gives |k| = 4.
The line meets the circle twice when 64 − 4k² > 0, that is −4 < k < 4.
The mistake to watch for
A frequent slip is stopping at one value of k.
Mistaken answer: k² = 16, so k = 4
The student took only the positive square root and did not notice the second tangent.
The correction is to write ± every time a square root is taken: k = ±4. Sketch the circle and the line y = x + k to see why: there is a tangent on each side of the circle. Another slip is setting the discriminant greater than zero, which finds a range of k, not the tangent values.
Check yourself
1. Find k so that y = 2x + k is a tangent to x² + y² = 5.
Show answer
x² + (2x + k)² = 5 gives 5x² + 4kx + k² − 5 = 0. The discriminant is 16k² − 20(k² − 5) = 100 − 4k². Set to 0: k² = 25.
k = 5 or k = −5. Check: |k| ÷ √5 = √5 gives |k| = 5. ✓
2. Find m so that y = mx + 4 is a tangent to x² + y² = 4.
Show answer
x² + (mx + 4)² = 4 gives (1 + m²)x² + 8mx + 12 = 0. The discriminant is 64m² − 48(1 + m²) = 16m² − 48. Set to 0: m² = 3.
m = √3 or m = −√3. Check: 4 ÷ √(1 + 3) = 2, the radius. ✓
3. For what values of k does y = x + k cut x² + y² = 8 at two distinct points?
Show answer
The discriminant is 64 − 4k². It must be greater than 0, so k² < 16.
−4 < k < 4.
Where this leads next
Next, reverse the direction in translating a geometric condition into a circle equation, then test the topic with the circle practice set. The line and circle intersection explorer shows the tangent case, and the quadratic structure explorer shows how the discriminant behaves as a coefficient changes.
When a student can do each step separately but loses the plan in a question with a letter in it, a teacher can practise planning with you in online one-to-one Additional Mathematics tuition.