Circle Properties

Circle theorems, line intersections, and the discriminant test

SPEC 3.2 • COORDINATE GEOMETRY

Angle in a Semicircle

Theorem

If AB is a diameter of a circle and P is any point on the circumference (not A or B), then angle APB = 90°

Why It Works

The gradient of AP × gradient of BP = −1 (perpendicular condition)

This property is essential for proving points lie on a circle or finding circle equations.

Proving Angle in Semicircle

If A = (−r, 0), B = (r, 0) are ends of a diameter, and P = (x, y) is on the circle x² + y² = r²:

Proof

Gradient of AP:

m_{1} = \frac{y - 0}{x - ( - r )} = \frac{y}{x + r}

Gradient of BP:

m_{2} = \frac{y - 0}{x - r} = \frac{y}{x - r}

Product:

m_{1} \times m_{2} = \frac{y}{x + r} \times \frac{y}{x - r} = \frac{y ^{2}}{x ^{2} - r ^{2}}

Since P is on circle: x² + y² = r², so y² = r² − x²

m_{1} \times m_{2} = \frac{r ^{2} - x ^{2}}{x ^{2} - r ^{2}} = \frac{- ( x ^{2} - r ^{2} )}{x ^{2} - r ^{2}} = - 1

✓

Application

Use this to find the equation of a circle when you know the diameter endpoints, or to prove that angle APB = 90°.

Circle-Line Intersections

To find where a line intersects a circle, substitute the line equation into the circle equation.

Method

Step 1: Rearrange line to make y (or x) the subject

Step 2: Substitute into circle equation

Step 3: Solve the resulting quadratic

Step 4: Find corresponding y (or x) values

Result

The quadratic gives 0, 1, or 2 solutions depending on whether the line misses, touches, or crosses the circle.

Worked Example01

Find the coordinates of the points where the line y = x + 2 intersects the circle x² + y² = 10.

1

Substitute y = x + 2 into x² + y² = 10:

•

x² + (x + 2)² = 10

2

Expand and simplify:

•

x² + x² + 4x + 4 = 10

•

2x² + 4x − 6 = 0

•

x² + 2x − 3 = 0

3

Factorise: (x + 3)(x − 1) = 0

•

x = −3 or x = 1

4

Find y values using y = x + 2:

•

When x = −3: y = −3 + 2 = −1

•

When x = 1: y = 1 + 2 = 3

Answer: (−3, −1) and (1, 3)

The Discriminant Test

When you substitute a line into a circle, you get a quadratic. The discriminant tells you about intersections:

Discriminant

det = b^{2} - 4 a c

DiscriminantMeaningGeometry

b^{2} - 4 a c > 0

Two distinct real rootsLine crosses circle at 2 points

b^{2} - 4 a c = 0

One repeated rootLine is tangent to circle

b^{2} - 4 a c < 0

No real rootsLine misses the circle

Key Use

Use the discriminant to prove a line is a tangent (set det = 0) or find conditions for intersection.

Visualising the Discriminant

Worked Example02

Prove that the line y = 2x + 5 is a tangent to the circle x² + y² = 5.

1

Substitute y = 2x + 5 into x² + y² = 5:

•

x² + (2x + 5)² = 5

2

Expand:

•

x² + 4x² + 20x + 25 = 5

•

5x² + 20x + 20 = 0

•

x² + 4x + 4 = 0

3

Calculate discriminant: a = 1, b = 4, c = 4

•

det = b² − 4ac = 16 − 16 = 0

4

Since det = 0, there is exactly one solution.

Therefore the line is a tangent to the circle. ✓

Worked Example03

Find the values of k for which the line y = x + k is a tangent to the circle x² + y² = 8.

1

Substitute y = x + k into x² + y² = 8:

•

x² + (x + k)² = 8

•

x² + x² + 2kx + k² = 8

•

2x² + 2kx + (k² − 8) = 0

2

For tangency, discriminant = 0:

•

a = 2, b = 2k, c = k² − 8

•

det = (2k)² − 4(2)(k² − 8) = 0

•

4k² − 8k² + 64 = 0

•

−4k² + 64 = 0

•

k² = 16

Answer: k = 4 or k = −4

Proving Points Lie on a Circle

To verify that a point lies on a circle, substitute its coordinates into the circle equation.

Method

Step 1: Take the point (a, b) you want to test

Step 2: Substitute x = a, y = b into circle equation

Step 3: If LHS = RHS, the point lies on the circle

Example

Does (3, 4) lie on the circle x² + y² = 25?

Check: 3² + 4² = 9 + 16 = 25 ✓ Yes!

Common Question Type

You may be given 3 points and asked to show they all lie on a circle, then find its equation.

Worked Example04

Show that the points A(0, 0), B(6, 0), and C(6, 8) all lie on the circle with equation (x − 3)² + (y − 4)² = 25.

A:

Point (0, 0):

•

(0 − 3)² + (0 − 4)² = 9 + 16 = 25 ✓

B:

Point (6, 0):

•

(6 − 3)² + (0 − 4)² = 9 + 16 = 25 ✓

C:

Point (6, 8):

•

(6 − 3)² + (8 − 4)² = 9 + 16 = 25 ✓

All three points satisfy the equation, so they lie on the circle.

Circle with Given Diameter

If you know the endpoints of a diameter, you can find the circle equation directly.

Method

Given: Diameter endpoints A(x₁, y₁) and B(x₂, y₂)

Centre: Midpoint of AB =

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Radius: Half of |AB| =

\frac{1}{2} (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Using Angle in Semicircle

If AB is a diameter and P is on the circle, then angle APB = 90°.

Conversely: if angle APB = 90°, then P lies on the circle with diameter AB.

Worked Example05

Points A(−2, 1) and B(4, 5) are the ends of a diameter. Find the equation of the circle.

1

Find the centre (midpoint of AB):

•

Centre = (\frac{- 2 + 4}{2}, \frac{1 + 5}{2}) = (1, 3)

2

Find the radius (half the diameter length):

•

|AB| =

(4 - (- 2))^{2} + (5 - 1)^{2} = 36 + 16 = 52

•

Radius =

\frac{52}{2}

, so r² =

\frac{52}{4} = 13

3

Write the equation:

Answer: (x − 1)² + (y − 3)² = 13

Combined Problems

Exam questions often combine circle-line intersections with other skills:

Common Combinations

Find intersection points, then calculate distance between them

Find intersection points, then calculate area of triangle with origin

Find the midpoint of a chord

Find the range of k for two intersections / tangency / no intersection

Strategy

First find the intersection points, then apply the relevant formula (distance, area, midpoint) to those coordinates.

Worked Example06

The line y = x + 2 intersects the circle x² + y² = 10 at points P and Q. Find the length PQ.

1

From Example 1, the intersection points are:

•

P = (−3, −1) and Q = (1, 3)

2

Use the distance formula:

•

|PQ| =

(1 - (- 3))^{2} + (3 - (- 1))^{2}

•

|PQ| =

4^{2} + 4^{2}

•

|PQ| =

16 + 16 = 32

Answer: |PQ| = 4√2

Worked Example07

Find the range of values of k for which the line y = x + k intersects the circle x² + y² = 8 at two distinct points.

1

From Example 3, we found the quadratic:

•

2x² + 2kx + (k² − 8) = 0

2

For two distinct points, discriminant > 0:

•

Δ = (2k)² − 4(2)(k² − 8) > 0

•

4k² − 8k² + 64 > 0

•

−4k² + 64 > 0

•

64 > 4k²

•

k² < 16

Answer: −4 < k < 4

Practice

Question 1: Find the values of c for which the line y = 3x + c is a tangent to the circle x² + y² = 10.

Substitute: x² + (3x + c)² = 10

x² + 9x² + 6cx + c² = 10

10x² + 6cx + (c² − 10) = 0

For tangency, det = 0:

(6c)² − 4(10)(c² − 10) = 0

36c² − 40c² + 400 = 0

−4c² + 400 = 0

c² = 100

Answer: c = 10 or c = −10

Practice

Question 2: Points A(−4, 0) and B(4, 0) are the ends of a diameter of a circle. Point P(0, 4) lies on the circle. Verify that angle APB = 90° using gradients.

Gradient of AP:

m_{A P} = \frac{4 - 0}{0 - ( - 4 )} = \frac{4}{4} = 1

Gradient of BP:

m_{BP} = \frac{4 - 0}{0 - 4} = \frac{4}{- 4} = - 1

Check product:

m₁ × m₂ = 1 × (−1) = −1

Since the product = −1, AP ⊥ BP, so angle APB = 90° ✓

Practice

Question 3: The line y = 2x + k intersects the circle (x − 2)² + (y − 1)² = 20 at two points. Find the range of values of k.

Substitute y = 2x + k:

(x − 2)² + (2x + k − 1)² = 20

x² − 4x + 4 + 4x² + 4x(k−1) + (k−1)² = 20

5x² + (4k − 8)x + (4 + k² − 2k + 1 − 20) = 0

5x² + (4k − 8)x + (k² − 2k − 15) = 0

For two points, det > 0:

(4k − 8)² − 4(5)(k² − 2k − 15) > 0

16k² − 64k + 64 − 20k² + 40k + 300 > 0

−4k² − 24k + 364 > 0

k² + 6k − 91 < 0

(k + 13)(k − 7) < 0

Answer: −13 < k < 7

Summary

Property/MethodKey Points

Angle in semicircleIf AB is diameter, angle APB = 90° for P on circle

Circle-line intersectionSubstitute line into circle, solve quadratic

det > 0Two intersection points

det = 0Line is tangent (one point)

det < 0No intersection

Proving point on circleSubstitute coordinates, check equation satisfied

Exam Strategy

When asked to prove tangency, use discriminant = 0. When finding range of k, set up the discriminant inequality.