Circles Meeting Lines, and Tangents

Find where a line meets a circle, use the discriminant to tell a chord from a tangent or a miss, and find the equation of a tangent.

In the exam for IGCSE Additional Maths 0606 · 7 real past-paper questions below

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Three ways a line meets a circle

Draw a straight line across a circle. It can cut through in two places, just touch in one place, or miss completely.

Two points

The line cuts the circle. The part inside is a chord.

One point or none

One point: the line just touches. It is a tangent.

No points: the line misses.

Key idea

A wheel rolling on a flat road shows a real tangent. The road touches the round tyre at one point only. The line from the centre of the wheel to that point stands straight up, at a right angle to the road.

To find the meeting points, both equations must be true at once. So put the line into the circle. You get a quadratic, and each solution is a meeting point.

Remember from the discriminant lesson: for $a x^{2} + b x + c = 0$ , the value of $b^{2} - 4 a c$ tells you how many solutions there are.

Formula

b^{2} - 4 a c

$a, b, c$ — the three numbers in the quadratic you get after putting the line into the circle and collecting everything on one side, equal to 0.

You see	You do
$b^{2} - 4 a c > 0$	Two points: the line is a chord
$b^{2} - 4 a c = 0$	One point: the line is a tangent
$b^{2} - 4 a c < 0$	No points: the line misses

Key idea

Substitute the line into the circle. The quadratic you get answers every question: where they meet, and how many times.

Finding the meeting points

To find the points, solve the quadratic, then put each x back into the line to get its y.

Steps

1
1. Make x or y the subject of the line. Pick the one that avoids fractions.
2. Substitute into the circle equation.
3. Expand, collect on one side and solve the quadratic.
4. Put each answer into the line to find the other coordinate. Pair them up.

Worked example

1
Find where the line $y = x + 1$ meets the circle $(x - 2)^{2} + (y - 1)^{2} = 10$ .
2
$(x - 2)^{2} + (x + 1 - 1)^{2} = 10$ — Replace y with x + 1.
3
$x^{2} - 4 x + 4 + x^{2} = 10$ — Expand the bracket. The second bracket is just x.
4
$2 x^{2} - 4 x - 6 = 0$ — Collect everything on one side.
5
$x^{2} - 2 x - 3 = 0$ — Divide by 2 to make it easier.
6
$(x - 3) (x + 1) = 0 \Rightarrow x = 3 or x = - 1$ — Factorise.
7
$(3, 4) and (- 1, 0)$ — Use y = x + 1 for each x.

Where a circle meets an axis is the same idea. The y-axis is the line $x = 0$ , and the x-axis is the line $y = 0$ .

Worked example

1
Find, in exact form, where the circle $(x - 3)^{2} + (y + 1)^{2} = 20$ meets the y-axis.
2
$(0 - 3)^{2} + (y + 1)^{2} = 20$ — On the y-axis, x = 0.
3
$(y + 1)^{2} = 11$ — 20 − 9 = 11.
4
$y + 1 = \pm 11$ — Square root both sides. Keep both signs.
5
$(0, - 1 + 11) and (0, - 1 - 11)$ — Exact form: leave the surd.

Common mistake

Stopping after you find the x values. A point needs both coordinates. Put each x into the line, not into the circle, so you do not pick up extra wrong points.

Common mistake

Losing the minus answer when you square root. $(y + 1)^{2} = 11$ gives two values of y.

Key idea

Line into circle, solve, then back into the line. Two answers mean two points: write them as a pair of coordinates each.

Show that a line is a tangent

A common question says “Show that the line is a tangent to the circle”. You must prove there is only one meeting point.

Worked example

1
Show that $y = 14 - 2 x$ is a tangent to $(x - 1)^{2} + (y - 2)^{2} = 20$ , and find the point where it touches.
2
$(x - 1)^{2} + (12 - 2 x)^{2} = 20$ — Replace y. Then 14 − 2x − 2 = 12 − 2x.
3
$x^{2} - 2 x + 1 + 144 - 48 x + 4 x^{2} = 20$ — Expand both brackets.
4
$5 x^{2} - 50 x + 125 = 0$ — Collect everything on one side.
5
$x^{2} - 10 x + 25 = 0$ — Divide by 5.
6
$b^{2} - 4 a c = (- 10)^{2} - 4 (1) (25) = 0$ — The discriminant is 0.
7
$One repeated root, so the line is a tangent$ — Write this conclusion. It earns the last mark.
8
$(x - 5)^{2} = 0 \Rightarrow x = 5, y = 4$ — The point of contact is (5, 4).

Rule

A “show that” answer needs a conclusion sentence. Write: the discriminant is 0, so the line is a tangent.

You may also factorise and show a repeated root, as in the last line. Either method earns the marks.

Common mistake

Working out $b^{2} - 4 a c$ before everything is on one side. The quadratic must equal 0 first, or a, b and c are wrong.

Key idea

Tangent means exactly one meeting point. In algebra, that is a discriminant of 0.

An unknown in the line

Sometimes the line has a letter in it, like $y = 2 x + k$ . The question asks for the values of k that make it a tangent. Keep k in the working and set the discriminant to 0.

Worked example

1
The line $y = 2 x + k$ is a tangent to $(x - 1)^{2} + (y + 2)^{2} = 5$ . Find the possible values of k.
2
$(x - 1)^{2} + (2 x + k + 2)^{2} = 5$ — Replace y with 2x + k.
3
$x^{2} - 2 x + 1 + 4 x^{2} + 4 (k + 2) x + (k + 2)^{2} = 5$ — Expand both brackets. Treat $k + 2$ as one number: $(2 x + m)^{2} = 4 x^{2} + 4 m x + m^{2}$ with $m = k + 2$ .
4
$5 x^{2} + (4 k + 6) x + (k^{2} + 4 k) = 0$ — Collect the $x^{2}$ terms, the x terms and the rest: $(k + 2)^{2} + 1 - 5 = k^{2} + 4 k$ . Keep k inside b and c.
5
$(4 k + 6)^{2} - 20 (k^{2} + 4 k) = 0$ — Tangent, so the discriminant is 0. Here a = 5.
6
$- 4 k^{2} - 32 k + 36 = 0$ — $(4 k + 6)^{2} = 16 k^{2} + 48 k + 36$ . Then subtract $20 k^{2} + 80 k$ .
7
$k^{2} + 8 k - 9 = 0$ — Divide by −4.
8
$(k - 1) (k + 9) = 0 \Rightarrow k = 1 or k = - 9$ — Two tangents with gradient 2, one on each side of the circle.

Now the question can go one step further. Between $k = - 9$ and $k = 1$ , the line sits between the two tangents, so it crosses the circle twice: it is a chord. Outside that range, it misses.

Rule

For a chord, solve discriminant > 0. For a miss, solve discriminant < 0. For a tangent, solve discriminant = 0.

Common mistake

Forgetting k in the c term. Every number without an x goes into c, including $k^{2}$ and $4 k$ .

Key idea

A letter in the line does not change the method. Substitute, collect, then use the discriminant on the letter.

Tangent at a point on the circle

If you know the point where the tangent touches, there is a faster way, and no calculus is needed. The tangent meets the radius at a right angle.

Rule

The tangent is perpendicular to the radius. So the gradient of the tangent is the negative reciprocal of the gradient of the radius.

Formula

m_{tangent} = - \frac{1}{m _{radius}}

$m_{radius}$ — the gradient from the centre to the point on the circle.
$m_{tangent}$ — flip the radius gradient upside down and change its sign.

Worked example

1
Find the equation of the tangent to $(x + 1)^{2} + (y - 3)^{2} = 25$ at the point $(2, 7)$ .
2
$Centre (- 1, 3)$ — Read the centre from the equation.
3
$m_{radius} = \frac{7 - 3}{2 - ( - 1 )} = \frac{4}{3}$ — Gradient from the centre to (2, 7).
4
$m_{tangent} = - \frac{3}{4}$ — Negative reciprocal.
5
$y - 7 = - \frac{3}{4} (x - 2)$ — Use the point (2, 7), not the centre.
6
$4 y - 28 = - 3 x + 6$ — Multiply both sides by 4.
7
$3 x + 4 y = 34$ — Integer form, as questions often ask.

Check: $3 (2) + 4 (7) = 34$ . The point is on your line. The mark scheme gives a mark for the centre, one for the radius gradient, one for using the perpendicular gradient at the point, and one for the answer.

The same right angle gives the length of a tangent from a point outside. The centre, the outside point and the touching point make a right-angled triangle.

Worked example

1
Find the length of the tangent from $P (8, 7)$ to $(x - 2)^{2} + (y - 1)^{2} = 20$ .
2
$d^{2} = (8 - 2)^{2} + (7 - 1)^{2} = 72$ — Distance squared from the centre (2, 1) to P.
3
$ℓ^{2} = 72 - 20 = 52$ — Pythagoras: the radius squared is 20.
4
$ℓ = 52 = 213$ — Exact form.

Common mistake

Using the radius gradient as the tangent gradient. The tangent is at a right angle to the radius.

Common mistake

Putting the centre into $y - y_{1} = m (x - x_{1})$ . The tangent passes through the point on the circle, not the centre.

Key idea

At a known point, use the right angle: radius gradient, flip and change sign, then the line through the point.

Which method? Read the question

The wording tells you which tool to use.

You see	You do
“Find the points where the line meets the circle”	Substitute the line, solve, then y from the line
“Where the circle meets the x-axis” or “y-axis”	Put y = 0 or x = 0 into the circle
“Show that the line is a tangent”	Substitute, then $b^{2} - 4 a c = 0$ and a conclusion
A letter in the line, “is a tangent”	Discriminant = 0, solve for the letter
“Meets the circle at two points”	Discriminant > 0
“Does not meet the circle”	Discriminant < 0
“Tangent at the point (p, q)”	Radius gradient, negative reciprocal, line through (p, q)
Length of a tangent from a point	Pythagoras: distance squared minus $r^{2}$

Common mistake

Rounding surds on 0606 Paper 1. There is no calculator, so give points like $(0, - 1 + 11)$ in exact form.

Key idea

Two tools do every question: substitute and use the discriminant, or use the right angle between tangent and radius.

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