Course Notes Practice Past Paper Archive

Intersections of Lines and Circles0%

Circles · Topic 7 of 7

Intersections of Lines and Circles

Video lesson · from 29:073 worked examples

One lesson video covers all of Circles, so it opens at 29:07 for this topic — not from the beginning.

Theory

We have previously studied the intersection of lines and curves. Similarly, lines and circles can intersect at two points, one point (tangent), or no points.

Two intersections

One intersection
i.e. tangent

No intersections

We can use substitution to find out how many times a line and a circle touch.

⚠️ Common Examiner Traps

Answer the question that was asked: a common waste of time is finding the centre and radius when the question does not require them. Substituting the line into the circle is the method — the centre and radius are not the answer.
Substitute, then use the discriminant: two intersections need $b^2-4ac \gt 0$ , a tangent needs $= 0$ , and no intersection needs $\lt 0$ . State which you are testing.
Rearrange to $ax^2+bx+c=0$ first: after substituting you must collect everything on one side before reading off $a$ , $b$ and $c$ .
Give the points if asked: solving gives $x$ only. Substitute back into the line — much easier than the circle — to get each $y$ , and write the coordinates.

Worked examples

Example 1

Find the points where the line with equation $y=3x$ intersects the circle with equation $x^2+y^2=20$ .

Substitute $y=3x$ into the circle equation:

x^2 + (3x)^2 = 20

x^2 + 9x^2 = 20

10x^2 = 20

x^2 = 2

x = \pm \sqrt{2}

Now find the corresponding $y$ values using $y=3x$ :

For $x = \sqrt{2}$ , $y = 3\sqrt{2}$ .

For $x = -\sqrt{2}$ , $y = -3\sqrt{2}$ .

The points of intersection are $(\sqrt{2}, 3\sqrt{2})$ and $(-\sqrt{2}, -3\sqrt{2})$ .

Example 2

Find the points where the line with equation $y=2x+6$ and circle with equation $x^2+y^2+2x+2y-8=0$ intersect.

Substitute $y = 2x+6$ into the circle equation:

x^2 + (2x+6)^2 + 2x + 2(2x+6) - 8 = 0

Expand the brackets:

x^2 + (4x^2 + 24x + 36) + 2x + (4x + 12) - 8 = 0

Simplify:

5x^2 + 30x + 40 = 0

Divide by 5:

x^2 + 6x + 8 = 0

Factorise:

(x+2)(x+4) = 0

Roots: $x = -2$ and $x = -4$ .

Find corresponding $y$ values:

For $x = -2$ , $y = 2(-2)+6 = 2$ .

For $x = -4$ , $y = 2(-4)+6 = -2$ .

The points of intersection are $(-2, 2)$ and $(-4, -2)$ .

Example 3

Show that the line $x-3y=5$ is a tangent to the circle with equation $x^2+y^2-6x+8y+15=0$ .

Rearrange the line equation to substitute: $x = 3y+5$ .

Substitute into the circle equation:

(3y+5)^2 + y^2 - 6(3y+5) + 8y + 15 = 0

Expand:

(9y^2+30y+25) + y^2 - (18y+30) + 8y + 15 = 0

Simplify:

10y^2 + 20y + 10 = 0

Divide by 10:

y^2 + 2y + 1 = 0

Calculate the discriminant for this quadratic ( $a=1, b=2, c=1$ ):

b^2 - 4ac = (2)^2 - 4(1)(1) = 4 - 4 = 0

Since the discriminant is 0, the equation has one repeated real root. Therefore, the line touches the circle at exactly one point, meaning it is a tangent.