Lines and circles

AS · P1 · 18 min

A straight line and a circle can meet in two points, touch at one point, or miss each other completely. Paper 1 circle questions almost always end here: find where a line cuts a circle, show a line is a tangent, find the values of a constant for which a line is a tangent, or work with the tangents from a point outside the circle. There are two ways in, algebra (substitute and use the discriminant) and geometry (compare the distance from the centre with the radius), and the strongest answers use whichever is quicker.

Three possibilities

Picture a line sliding across a circle. Far away it misses; as it comes in it first touches the circle at one point; then it cuts the circle at two points, forming a chord.

x^2 + y^2 = 9 y = x + 1 y = x + 3*sqrt(2) y = x + 6

The algebra mirrors the picture. Substituting the line into the circle gives a quadratic, and the number of real roots is the number of meeting points. The geometry says the same thing with distances: let $d$ be the shortest distance from the centre to the line.

Key result

Discriminant of the quadratic	Distance from centre to line	The line
$b^2 - 4ac > 0$	$d < r$	cuts the circle at two points (a chord)
$b^2 - 4ac = 0$	$d = r$	is a tangent (touches at one point)
$b^2 - 4ac < 0$	$d > r$	does not meet the circle

The substitution itself is the routine from Simultaneous linear and quadratic equations: make $x$ or $y$ the subject of the linear equation, substitute into the circle, expand, and collect into $ax^2 + bx + c = 0$ . This note concentrates on what to do with the result.

Chords

When a line cuts a circle at $A$ and $B$ , the segment $AB$ is a chord. Two facts make chord questions quick:

The perpendicular from the centre to a chord bisects the chord. So the midpoint $M$ of $AB$ is the foot of the perpendicular from the centre $C$ .
Triangle $CMA$ is right-angled at $M$ , so

\left(\tfrac{1}{2}AB\right)^2 + CM^2 = r^2

This gives the chord length from the distance $CM$ without finding $A$ and $B$ , which helps when the intersection points are surds.

Tangents

A line is a tangent when it meets the circle at exactly one point. There are three standard situations.

Showing a given line is a tangent

Substitute and show the quadratic has a repeated root (it factorises as a perfect square, or $b^2 - 4ac = 0$ ). The repeated root gives the point of contact. Always finish with the sentence "repeated root, so the line is a tangent".

A tangent with an unknown constant

When the line contains a constant $k$ , the quadratic has $k$ in its coefficients. Setting $b^2 - 4ac = 0$ gives an equation for $k$ , usually with two solutions: two parallel tangents, one on each side of the circle. Replacing $= 0$ by $> 0$ or $< 0$ gives the values of $k$ for which the line cuts or misses the circle, solved as a quadratic inequality.

The geometric alternative: a tangent is at distance $r$ from the centre. Find the foot of the perpendicular from the centre to the line in terms of $k$ and set its distance equal to $r$ . Both methods give the same answer.

Tangents from a point outside the circle

From a point $T$ outside a circle there are exactly two tangents, touching at $P$ and $Q$ .

The radius to each point of contact is perpendicular to the tangent, so triangles $CPT$ and $CQT$ are right-angled.
By Pythagoras, the length of each tangent is $TP = TQ = \sqrt{CT^2 - r^2}$ .
The figure $CPTQ$ is a kite, symmetrical about the line $CT$ .

To find the tangents themselves, write the line through $T$ with unknown gradient $m$ , substitute into the circle and set the discriminant to zero. The resulting quadratic in $m$ gives the two gradients.

x^2 + y^2 = 10 y = 3x - 10 y = -(1/3)x + 10/3 (4, 2) (3, -1) (1, 3) (0, 0) -- (3, -1) (0, 0) -- (1, 3)

The two tangents from $T(4, 2)$ to $x^2 + y^2 = 10$ touch at $(3, -1)$ and $(1, 3)$ ; each tangent has length $\sqrt{20 - 10} = \sqrt{10}$ .

Line and circle

Rearrange the line to make $x$ or $y$ the subject (choose to avoid fractions).
Substitute into the circle, bracketing the expression, and expand.
Collect into $ax^2 + bx + c = 0$ ; divide by any common factor.
Two distinct roots: find both points (use the line to get the other coordinate). Repeated root: tangent, find the contact point. Unknown constant: use $b^2 - 4ac$ with $=$ , $>$ or $<$ .
For chords, tangent lengths and shortest distances, look for a right-angled triangle with the radius as one side.

Worked examples

Intersection points and the length of a chord

The line $y = x + 2$ meets the circle $x^2 + y^2 = 10$ at $A$ and $B$ . Find the coordinates of $A$ and $B$ , and show that the perpendicular from the origin to $AB$ passes through the midpoint of $AB$ .

Solution

\begin{aligned} x^2 + (x + 2)^2 &= 10 \\ 2x^2 + 4x - 6 &= 0 \\ x^2 + 2x - 3 &= 0 \\ (x + 3)(x - 1) &= 0 \end{aligned}

$x = -3$ gives $y = -1$ ; $x = 1$ gives $y = 3$ . So $A(-3, -1)$ and $B(1, 3)$ .

$AB = \sqrt{4^2 + 4^2} = 4\sqrt{2}$ . Midpoint $M = (-1, 1)$ . The line $OM$ has gradient $-1$ and $AB$ has gradient $1$ , so $OM \perp AB$ : the perpendicular from the centre meets $AB$ at its midpoint, as the chord property says.

Showing a line is a tangent

Show that the line $x + 2y = 7$ is a tangent to the circle $(x - 1)^2 + (y + 2)^2 = 20$ , and find the point of contact.

Solution

From the line, $x = 7 - 2y$ , so $x - 1 = 6 - 2y$ :

\begin{aligned} (6 - 2y)^2 + (y + 2)^2 &= 20 \\ 36 - 24y + 4y^2 + y^2 + 4y + 4 &= 20 \\ 5y^2 - 20y + 20 &= 0 \\ y^2 - 4y + 4 &= 0 \\ (y - 2)^2 &= 0 \end{aligned}

There is a repeated root, so the line meets the circle at exactly one point and is a tangent. $y = 2$ gives $x = 7 - 4 = 3$ : the point of contact is $(3, 2)$ .

Values of a constant for a tangent

(a) Find the values of $k$ for which the line $y = 2x + k$ is a tangent to the circle $(x - 1)^2 + (y - 3)^2 = 5$ .

(b) State the set of values of $k$ for which the line meets the circle at two distinct points.

Solution

(a) Substitute $y = 2x + k$ , so $y - 3 = 2x + (k - 3)$ :

\begin{aligned} (x - 1)^2 + (2x + k - 3)^2 &= 5 \\ x^2 - 2x + 1 + 4x^2 + 4(k - 3)x + (k - 3)^2 - 5 &= 0 \\ 5x^2 + (4k - 14)x + (k - 3)^2 - 4 &= 0 \end{aligned}

Tangent means $b^2 - 4ac = 0$ :

\begin{aligned} (4k - 14)^2 - 20\left[(k - 3)^2 - 4\right] &= 0 \\ 16k^2 - 112k + 196 - 20k^2 + 120k - 100 &= 0 \\ -4k^2 + 8k + 96 &= 0 \\ k^2 - 2k - 24 &= 0 \\ (k - 6)(k + 4) &= 0 \end{aligned}

So $k = 6$ or $k = -4$ .

(b) Two points means $b^2 - 4ac > 0$ , i.e. $-4(k^2 - 2k - 24) > 0$ , so $k^2 - 2k - 24 < 0$ and $-4 < k < 6$ . This makes sense: the two tangents are the boundary lines, and every parallel line between them cuts the circle.

A chord length from the distance to the centre

The line $y = x - 1$ meets the circle $x^2 + y^2 - 4x + 6y - 12 = 0$ at $A$ and $B$ . Find the exact length of $AB$ .

Solution

Completing the square: $(x - 2)^2 + (y + 3)^2 = 25$ . Centre $C(2, -3)$ , $r = 5$ .

The perpendicular from $C$ to the line has gradient $-1$ : $y + 3 = -(x - 2)$ , i.e. $y = -x - 1$ . It meets $y = x - 1$ where $x - 1 = -x - 1$ , so $x = 0$ , $y = -1$ . The foot is $M(0, -1)$ , the midpoint of $AB$ .

CM^2 = 2^2 + 2^2 = 8, \qquad \left(\tfrac{1}{2}AB\right)^2 = r^2 - CM^2 = 25 - 8 = 17

So $AB = 2\sqrt{17}$ .

(Substituting instead gives $2x^2 - 17 = 0$ , so $x = \pm\sqrt{\tfrac{17}{2}}$ : the same length, with more surd work.)

Tangents from an external point

The point $T$ is $(4, 2)$ and the circle is $x^2 + y^2 = 10$ .

(a) Find the length of a tangent from $T$ to the circle.

(b) Find the equations of the two tangents from $T$ , and their points of contact.

Solution

(a) Centre $O(0, 0)$ , $r^2 = 10$ , $OT^2 = 16 + 4 = 20$ . The tangent is perpendicular to the radius, so

\text{tangent length} = \sqrt{OT^2 - r^2} = \sqrt{20 - 10} = \sqrt{10}

(b) A line through $T$ with gradient $m$ is $y = mx + 2 - 4m$ . Substitute into the circle:

x^2 + (mx + 2 - 4m)^2 = 10 \quad\Rightarrow\quad (1 + m^2)x^2 + 2m(2 - 4m)x + (2 - 4m)^2 - 10 = 0

Tangent: $b^2 - 4ac = 0$ . Dividing by $4$ :

\begin{aligned} m^2(2 - 4m)^2 - (1 + m^2)\left[(2 - 4m)^2 - 10\right] &= 0 \\ -(2 - 4m)^2 + 10(1 + m^2) &= 0 \\ -4 + 16m - 16m^2 + 10 + 10m^2 &= 0 \\ 3m^2 - 8m - 3 &= 0 \\ (3m + 1)(m - 3) &= 0 \end{aligned}

$m = 3$ : the tangent is $y = 3x - 10$ . Then $x^2 + (3x - 10)^2 = 10$ gives $10(x - 3)^2 = 0$ , contact $(3, -1)$ .

$m = -\tfrac{1}{3}$ : the tangent is $y = -\tfrac{1}{3}x + \tfrac{10}{3}$ , i.e. $x + 3y = 10$ . Then $x = 10 - 3y$ gives $10(y - 3)^2 = 0$ , contact $(1, 3)$ .

Check: the distance from $T$ to each contact point is $\sqrt{1 + 9} = \sqrt{10}$ , matching (a).

Tangents at the ends of a chord

The circle $x^2 + y^2 - 6x - 2y - 15 = 0$ has centre $C$ . The line $x + 7y = 35$ meets the circle at $A$ and $B$ .

(a) Find the coordinates of $A$ and $B$ .

(b) The tangents at $A$ and $B$ meet at $T$ . Find $T$ .

Solution

(a) $(x - 3)^2 + (y - 1)^2 = 25$ , so $C = (3, 1)$ , $r = 5$ . Substitute $x = 35 - 7y$ :

\begin{aligned} (35 - 7y)^2 + y^2 - 6(35 - 7y) - 2y - 15 &= 0 \\ 50y^2 - 450y + 1000 &= 0 \\ y^2 - 9y + 20 &= 0 \\ (y - 4)(y - 5) &= 0 \end{aligned}

$y = 4$ gives $x = 7$ ; $y = 5$ gives $x = 0$ . So $A(0, 5)$ and $B(7, 4)$ .

(b) The radius $CA$ has gradient $\dfrac{5 - 1}{0 - 3} = -\dfrac{4}{3}$ , so the tangent at $A$ has gradient $\tfrac{3}{4}$ : $y - 5 = \tfrac{3}{4}x$ , i.e. $3x - 4y = -20$ .

The radius $CB$ has gradient $\dfrac{4 - 1}{7 - 3} = \dfrac{3}{4}$ , so the tangent at $B$ has gradient $-\tfrac{4}{3}$ : $y - 4 = -\tfrac{4}{3}(x - 7)$ , i.e. $4x + 3y = 40$ .

Solving: $9x - 12y = -60$ and $16x + 12y = 160$ add to $25x = 100$ , so $x = 4$ and $y = 8$ . $T = (4, 8)$ .

(c) $TA = \sqrt{16 + 9} = 5$ and $TB = \sqrt{9 + 16} = 5$ (equal tangents). Each of triangles $CAT$ and $CBT$ is right-angled with legs $5$ and $5$ :

\text{Area} = 2 \times \tfrac{1}{2} \times 5 \times 5 = 25

Since $CA = AT = TB = BC = 5$ and the angles at $A$ and $B$ are right angles, $CATB$ is in fact a square.

A line that misses the circle

Show that the line $x + 2y = 15$ does not meet the circle $(x - 1)^2 + (y - 2)^2 = 5$ , and find the shortest distance between them.

Solution

Substitute $x = 15 - 2y$ , so $x - 1 = 14 - 2y$ :

(14 - 2y)^2 + (y - 2)^2 = 5 \quad\Rightarrow\quad 5y^2 - 60y + 195 = 0 \quad\Rightarrow\quad y^2 - 12y + 39 = 0

$b^2 - 4ac = 144 - 156 = -12 < 0$ : no real roots, so the line does not meet the circle.

The perpendicular from the centre $(1, 2)$ has gradient $2$ : $y = 2x$ . It meets the line where $x + 4x = 15$ , so the foot is $(3, 6)$ . The distance from the centre to the line is $\sqrt{2^2 + 4^2} = 2\sqrt{5}$ .

The closest point of the circle lies on this perpendicular, one radius from the centre, so the shortest distance is

2\sqrt{5} - \sqrt{5} = \sqrt{5}

Watch out

Substituting into the wrong bracket. With $(x - 1)^2 + (y - 3)^2 = 5$ and $y = 2x + k$ , the second bracket is $(2x + k - 3)^2$ , not $(2x + k)^2 - 3$ . Write $y - 3$ in terms of $x$ first.

Squaring a three-term bracket. $(2x + k - 3)^2 = 4x^2 + 4(k - 3)x + (k - 3)^2$ . Treat $k - 3$ as one term.

Forgetting to collect before using the discriminant. $a$ , $b$ and $c$ must be the coefficients of the fully collected quadratic $ax^2 + bx + c = 0$ .

Getting the other coordinate from the circle. From the circle, $y^2 = \ldots$ gives two values of $y$ for each $x$ , and one is not on the line. Use the line.

Confusing "meets" with "meets at two points". "Meets the circle" includes touching: $b^2 - 4ac \ge 0$ . "Two distinct points" is $> 0$ .

Exam tip

Typical marks. Substitution M1, correct three-term quadratic A1, discriminant condition M1, solving for $k$ A1 A1. Show the quadratic in $x$ clearly collected, because the examiner looks for it.
"Show that ... is a tangent". Factorise to a perfect square or show $b^2 - 4ac = 0$ , then write the conclusion. The point of contact is often the next part.
Choose the geometry when it is shorter. Chord lengths, tangent lengths and shortest distances are right-angled triangles with the radius as a side. State the property you are using.
Exact values. Leave answers like $2\sqrt{17}$ and $\sqrt{10}$ in surd form unless told otherwise.
Inequalities. For "two distinct points" or "does not meet", solve the quadratic inequality in $k$ with a sketch, and give the answer as a set of values.

Summary

Substitute the line into the circle to get $ax^2 + bx + c = 0$ . Two roots: chord; repeated root: tangent; no real roots: no intersection.
Equivalently, compare the distance $d$ from the centre to the line with $r$ .
The perpendicular from the centre bisects a chord: $\left(\tfrac{1}{2}AB\right)^2 = r^2 - d^2$ .
A tangent with unknown constant $k$ : set $b^2 - 4ac = 0$ ; there are usually two answers (two parallel tangents).
From an external point $T$ : two tangents, each of length $\sqrt{CT^2 - r^2}$ ; find their gradients with the discriminant in $m$ .
Tangents at the ends of a chord meet on the line through the centre perpendicular to the chord.
Shortest distance from a line to a circle it misses: (distance from centre to line) $-\ r$ .

Practice questions

Question

The line $y = 2x - 1$ meets the circle $x^2 + y^2 = 13$ at $A$ and $B$ . Find the coordinates of $A$ and $B$ .
Show that the line $x + y = 6$ does not meet the circle $x^2 + y^2 - 2x - 2y - 2 = 0$ .
Find the values of $k$ for which $y = x + k$ is a tangent to $x^2 + y^2 = 8$ , and the points of contact.
Find the set of values of $c$ for which the line $y = 2x + c$ meets the circle $x^2 + y^2 = 5$ at two distinct points.
Find the length of the chord cut from the circle $(x - 3)^2 + (y - 4)^2 = 25$ by the line $y = 1$ , using the distance from the centre.
Find the length of the tangents from the point $(7, 1)$ to the circle $x^2 + y^2 - 4x + 2y - 4 = 0$ .
Find the equations of the two tangents to $x^2 + y^2 = 20$ that have gradient $2$ , and their points of contact.
Find the shortest distance between the circle $(x + 1)^2 + (y - 2)^2 = 9$ and the line $3x + 4y = 30$ .
Tangents are drawn from $T(5, 5)$ to the circle $x^2 + y^2 = 10$ . (a) Find the gradients of the two tangents and their equations. (b) Find the points of contact and the area of the kite formed by $O$ , $T$ and the two points of contact.
The circle $x^2 + y^2 - 8x - 4y + 15 = 0$ has centre $C$ . (a) Show that the line $y = 2x - 1$ is a tangent to the circle and find the point of contact $P$ . (b) Find the equation of the other tangent to the circle that is parallel to $y = 2x - 1$ . (c) Find the exact distance between these two parallel tangents.

Answers

$x^2 + (2x - 1)^2 = 13$ gives $5x^2 - 4x - 12 = 0$ , $(5x + 6)(x - 2) = 0$ . $x = 2$ : $y = 3$ . $x = -\tfrac{6}{5}$ : $y = -\tfrac{17}{5}$ . Points $(2, 3)$ and $\left(-\tfrac{6}{5}, -\tfrac{17}{5}\right)$ .
$y = 6 - x$ : $x^2 + (6 - x)^2 - 2x - 2(6 - x) - 2 = 0$ simplifies to $2x^2 - 12x + 22 = 0$ , i.e. $x^2 - 6x + 11 = 0$ . $b^2 - 4ac = 36 - 44 = -8 < 0$ , so no real roots and no intersection. (Geometrically: centre $(1, 1)$ , $r = 2$ , and the foot of the perpendicular is $(3, 3)$ at distance $2\sqrt{2} > 2$ .)
$x^2 + (x + k)^2 = 8$ gives $2x^2 + 2kx + k^2 - 8 = 0$ . Tangent: $4k^2 - 8(k^2 - 8) = 0$ , so $k^2 = 16$ , $k = \pm 4$ . $k = 4$ : $2(x + 2)^2 = 0$ , contact $(-2, 2)$ . $k = -4$ : contact $(2, -2)$ .
$5x^2 + 4cx + c^2 - 5 = 0$ . Two points: $16c^2 - 20(c^2 - 5) > 0$ , so $100 - 4c^2 > 0$ , $c^2 < 25$ : $-5 < c < 5$ .
The distance from $(3, 4)$ to $y = 1$ is $3$ . Half-chord $= \sqrt{25 - 9} = 4$ , so the chord is $8$ . (The ends are $(-1, 1)$ and $(7, 1)$ .)
$(x - 2)^2 + (y + 1)^2 = 9$ : centre $(2, -1)$ , $r = 3$ . $CT^2 = 25 + 4 = 29$ . Tangent length $= \sqrt{29 - 9} = \sqrt{20} = 2\sqrt{5}$ .
$y = 2x + c$ : $5x^2 + 4cx + c^2 - 20 = 0$ . Tangent: $16c^2 - 20(c^2 - 20) = 0$ , so $c^2 = 100$ , $c = \pm 10$ . $y = 2x + 10$ touches at $(-4, 2)$ ; $y = 2x - 10$ touches at $(4, -2)$ .
Centre $(-1, 2)$ , $r = 3$ . The perpendicular from the centre has gradient $\tfrac{4}{3}$ : $4x - 3y = -10$ . With $3x + 4y = 30$ , this gives $x = 2$ , $y = 6$ . Distance from centre $= \sqrt{9 + 16} = 5$ . Shortest distance $= 5 - 3 = 2$ .
(a) $y = mx + 5 - 5m$ . Substituting gives $(1 + m^2)x^2 + 2m(5 - 5m)x + (5 - 5m)^2 - 10 = 0$ . Setting $b^2 - 4ac = 0$ and simplifying: $10(1 + m^2) = (5 - 5m)^2$ , so $15m^2 - 50m + 15 = 0$ , i.e. $3m^2 - 10m + 3 = 0$ , $(3m - 1)(m - 3) = 0$ . Tangents $y = 3x - 10$ and $y = \tfrac{1}{3}x + \tfrac{10}{3}$ (i.e. $x - 3y + 10 = 0$ ). (b) $m = 3$ : $10x^2 - 60x + 90 = 0$ , contact $(3, -1)$ . $m = \tfrac{1}{3}$ : $10x^2 + 20x + 10 = 0$ , contact $(-1, 3)$ . Tangent length $\sqrt{50 - 10} = \sqrt{40}$ . The kite is two right-angled triangles with legs $\sqrt{10}$ and $\sqrt{40}$ : area $= 2 \times \tfrac{1}{2}\sqrt{10}\sqrt{40} = 20$ .
(a) Substituting: $x^2 + (2x - 1)^2 - 8x - 4(2x - 1) + 15 = 0$ gives $5x^2 - 20x + 20 = 0$ , i.e. $5(x - 2)^2 = 0$ . Repeated root, so tangent; $P = (2, 3)$ . (b) $C = (4, 2)$ . The diametrically opposite point is $Q = 2C - P = (6, 1)$ , and the tangent there is parallel: $y - 1 = 2(x - 6)$ , i.e. $y = 2x - 11$ . (c) The distance between the parallel tangents is the diameter $PQ = \sqrt{16 + 4} = 2\sqrt{5}$ .

Three possibilities

Chords

Tangents

Showing a given line is a tangent

A tangent with an unknown constant

Tangents from a point outside the circle

Worked examples

Practice questions

Console