To find where a line meets a circle, substitute the line’s equation into the circle’s equation, solve the quadratic, then put each answer back into the line to find the other coordinate. The discriminant of that quadratic tells you whether there are two points, one point or none.
This skill sits in circle coordinate methods and follows the same pattern as the simultaneous equations in simultaneous linear and nonlinear models. It leads directly to tangents in later lessons.
Why does substitution give a quadratic?
A point on both graphs satisfies both equations at once. The line lets you write y in terms of x, and the circle contains x² and y². Replacing y by an expression in x produces terms in x², so you always get a quadratic in x.
A quadratic has at most two roots, which is why a line meets a circle at most twice.
How do you do it, step by step?
- Write the line with one variable as the subject, such as y = x + 2.
- Substitute into the circle, using brackets around the whole expression.
- Expand carefully, including the middle term of any bracket squared.
- Collect terms into the form ax² + bx + c = 0 and simplify by any common factor.
- Solve by factorising or the formula, and use the discriminant if only the number of points is needed.
- Find the second coordinate of each point from the line, then check it in the circle.
Worked example
The circle is x² + y² − 2x − 4y − 20 = 0 and the line is y = x + 2. Find the points of intersection and the length of the chord.
Step 1, substitute: x² + (x + 2)² − 2x − 4(x + 2) − 20 = 0.
Step 2, expand: x² + x² + 4x + 4 − 2x − 4x − 8 − 20 = 0.
Step 3, collect: 2x² − 2x − 24 = 0, so x² − x − 12 = 0.
Step 4, solve: (x − 4)(x + 3) = 0, so x = 4 or x = −3.
Step 5, find y from the line: x = 4 gives y = 6, and x = −3 gives y = −1. The points are (4, 6) and (−3, −1).
Step 6, chord length: √(7² + 7²) = √98 = 7√2.
Check: at (4, 6): 16 + 36 − 8 − 24 − 20 = 0 ✓. At (−3, −1): 9 + 1 + 6 + 4 − 20 = 0 ✓.
The mistake to watch for
The most frequent error is expanding (x + 2)² incorrectly.
Mistaken working: x² + (x + 2)² becomes x² + x² + 4
The student squared each term separately and lost the middle term 4x.
The correction is to write (x + 2)(x + 2) in full: x² + 2x + 2x + 4 = x² + 4x + 4. With the middle term kept, the quadratic comes out correctly, and in this example it factorises neatly. A quadratic that refuses to factorise is often a sign of a lost term.
Check yourself
1. Find the points where y = 3x meets x² + y² = 10.
Show answer
x² + 9x² = 10, so x² = 1 and x = ±1. Then y = 3x gives y = ±3.
Points (1, 3) and (−1, −3).
2. Show that y = x + 6 does not meet x² + y² = 8.
Show answer
x² + (x + 6)² = 8 gives 2x² + 12x + 28 = 0, so x² + 6x + 14 = 0. The discriminant is 36 − 56 = −20, which is negative, so there is no intersection.
3. Find the points where x + y = 3 meets x² + y² = 5.
Show answer
y = 3 − x, so x² + (3 − x)² = 5, which gives 2x² − 6x + 4 = 0 and x² − 3x + 2 = 0. Then x = 1 or x = 2.
Points (1, 2) and (2, 1). Check: 1 + 4 = 5 and 4 + 1 = 5. ✓
Where this leads next
Once substitution is reliable, move on to using a tangent perpendicular to the radius, then test the topic with the circle practice set. The line and circle intersection explorer shows the two, one or zero intersections as you move a line, and the non-calculator working trainer helps with the exact arithmetic.
Some students can repeat this example but cannot tell which variable to substitute when the line looks different. A teacher can spot that decision point in online one-to-one Additional Mathematics tuition.