01 · Pair the answers
Find where y = x + 1 meets x² + (y − 2)² = 13.
Hint
x² + (x − 1)² = 13.
Worked solution
x² − x − 6 = 0
x = −2 or 3
The points are (−2,−1) and (3,4).
Understand · explore · practise
Circle and line intersections
Solve circle and line equations simultaneously, use the discriminant to count intersections and find tangent conditions. Includes chord lengths and an extension on two intersecting circles.
Before you startCircle equations, simultaneous equations and the discriminant
01 / Substitution
Substitute the line into the circle, then pair the coordinates.
A common point must satisfy both equations. Express y in terms of x, substitute into the circle, solve the resulting quadratic and find the matching y for each x.
Keep exact roots. Checking both original equations avoids mismatching coordinates or retaining an algebra mistake.
Two exact intersection pointsWorked example
Circle: x² + (y − 1)² = 25
Line: y = x + 2
Replace y − 1 by x + 1.
x² + (x + 1)² = 25
2x² + 2x − 24 = 0
Divide by 2.
(x + 4)(x − 3) = 0
x = −4 or 3
Recover each y using the line.
(−4, −2) and (3, 5)
Check: 16 + 9 = 25 and 9 + 16 = 25.
02 / Axes and verticals
A vertical line is easiest to substitute directly.
On the x-axis set y = 0. On the y-axis set x = 0. For a vertical line x = k, substitute k and solve for y; no gradient is needed.
(x − 2)² + (y + 1)² = 10
On the x-axis: (x − 2)² = 9
Points: (−1,0), (5,0)
On the y-axis: (y + 1)² = 6
Points: (0,−1 ± √6)
The ± notation represents two separate coordinate pairs. State them separately when pairing with another ± expression could be ambiguous.
03 / Count intersections
The discriminant tells you how many real shared points exist.
A line and a circle of positive radius meet at two points, one tangent point, or no points. After substitution, use the discriminant Δ = b² − 4ac of the resulting quadratic.
Δ > 0: two intersections
Δ = 0: one tangent point
Δ < 0: no intersection
Here the circle is x² + y² = 25. A horizontal line y = k gives x² = 25 − k². The same argument works for a vertical line with x and y exchanged.
x² + y² = 25Equal axis scales
The circle x² + y² = 25 meets y = 3 at (−4,3) and (4,3). There are two intersections.
Watch a secant become a tangent
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Parameter ranges
Apply the discriminant to the substituted equation.
A parameter can control the line’s slope, the centre or the radius. Form the intersection quadratic first; its coefficients may depend on that parameter.
Which slopes cut the circle twice?Worked example
Circle: (x − 3)² + y² = 4
Line: y = kx
Substitute y = kx.
(1 + k²)x² − 6x + 5 = 0
Its leading coefficient is always positive.
Δ = 36 − 20(1 + k²)
= 16 − 20k²
Two distinct intersections require Δ > 0.
−2/√5 < k < 2/√5
At either endpoint the line is tangent. Outside this interval there is no intersection.
Remember that the circle must exist
For x² + 2x + y² = k and y = 3, the circle is (x + 1)² + y² = k + 1. Positive radius requires k > −1. Substitution gives (x + 1)² = k − 8, so a genuine circle with no intersection requires −1 < k < 8.
05 / Chord lengths
Intersections give endpoints you can use in geometry.
A chord joins two points on a circle. Once you have its endpoints, calculate its midpoint, length or perpendicular bisector using straight-line methods.
x² + y² = 25, y = 3
A = (−4,3), B = (4,3)
Midpoint (0,3), chord length 8
The centre (0,0) lies on the perpendicular bisector x = 0, but it is not the midpoint of this chord. A chord is a diameter only if it passes through the centre.
The distance from the centre to this chord is 3. Half the chord is √(5² − 3²) = 4 by Pythagoras, giving another route to length 8.
A diameter from simultaneous equations
(x − 1)² + (y − 2)² = 20 and y = 2x meet at (−1,−2) and (3,6). Their midpoint is (1,2), the centre, so this chord is a diameter. The line also passes through the centre directly.
06 / Two circles
Subtract two circle equations to remove both squared terms.
For two circles with different centres, subtraction gives a line containing every common point. Substitute that line into either circle to find the intersections. If they meet twice, it is the line of their common chord.
Find a common chord and a kite areaWorked example
C₁: x² + y² = 25
C₂: (x − 6)² + y² = 13
Their centres are O(0,0) and C(6,0).
Subtract C₁ from C₂:
−12x + 36 = −12 ⇒ x = 4
The common chord is vertical.
16 + y² = 25
P = (4,3), Q = (4,−3)
Both pairs satisfy both circle equations.
Area of kite OPCQ
= ½ × OC × PQ
= ½ × 6 × 6 = 18
Its diagonals OC and PQ are perpendicular. The centres lie on opposite sides of this chord.
With the same centre, unequal radii give no intersection. Identical circles have infinitely many common points; subtraction gives an identity rather than a chord line.
07 / Your turn
Count first when exact coordinates are not needed.
Use strict inequalities for two distinct points or no intersection; use equality for tangency.
02 · A vertical line
Find where x = 2 meets x² + y² = 25.
Hint
4 + y² = 25.
Worked solution
(2,√21) and (2,−√21).
03 · No meeting
Show that y = x + 8 does not meet x² + y² = 25.
Hint
Form the quadratic and use its discriminant.
Worked solution
2x² + 16x + 39 = 0
Δ = 256 − 312 = −56 < 0
There are no real common points.
04 · A tangent point
Show x + y = 10 is tangent to x² + y² = 50 and find the point.
Hint
Substitute y = 10 − x.
Worked solution
2x² − 20x + 50 = 0
2(x − 5)² = 0
There is one repeated root, x = 5, giving the contact point (5,5).
05 · A range of slopes
Find k for which y = kx meets (x − 4)² + y² = 4 twice.
Hint
(1 + k²)x² − 8x + 12 = 0.
Worked solution
Δ = 64 − 48(1 + k²)
= 16 − 48k² > 0
−1/√3 < k < 1/√3
06 · Chord and triangle
x² + y² = 25 meets y = −3 at A and B. Find AB and the area of OAB.
Hint
The endpoints have x = ±4; the perpendicular height from O is 3.
Worked solution
AB = 8
Area OAB = ½ × 8 × 3 = 12
07 · Two circles
Find the intersections of x² + y² = 25 and (x − 8)² + y² = 25.
Hint
Subtract the first equation from the second.
Worked solution
−16x + 64 = 0 ⇒ x = 4
y² = 9
The intersections are (4,3) and (4,−3).
08 / Recap
Each intersection satisfies both original equations.
- Substitute the line into the circle.
- Pair every x with its own y.
- Use Δ to count intersections and find parameter conditions.
- Check that a parameterised circle has positive radius.
- A diameter is a chord through the centre.
- Subtract circle equations to find their common-chord line.
Section 1 of 8 · Substitution