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Cambridge IGCSE · Topic 8.2
IGCSE Add Maths 0606 / Topic 8 / IGCSE

Chord, tangent or miss?

Substitute the line into the circle and you get a quadratic. Its two roots are the two points where the line cuts the circle, and its discriminant tells you at once whether the line is a chord, a tangent or misses altogether. The perpendicular from the centre gives a second test, and the midpoint and length of any chord.

Topic 8All coordinate geometry of the circle lessons →
The key idea

To find where a line meets a circle, make \(x\) or \(y\) the subject of the line, substitute into the circle equation and rearrange to a quadratic \(ax^{2} + bx + c = 0\). Then \(b^{2} - 4ac > 0\): two points, the line is a chord; \(b^{2} - 4ac = 0\): one point, the line is a tangent; \(b^{2} - 4ac < 0\): no points, the line misses the circle. Equivalently, compare the perpendicular distance \(d\) from the centre to the line with the radius \(r\): \(d < r\) chord, \(d = r\) tangent, \(d > r\) miss.

x y O C(2, 1) A(−1, 5) B(6, −2) M T(2, −4) F x + y = 4 Δ > 0: a chord AB y = −4 Δ = 0: a tangent at T x + y = 12 Δ < 0: misses the circle Circle: centre C, radius 5 CM ⊥ AB, so M is the midpoint of the chord. CF > 5 and CT = 5.
Fig. 1: the circle with centre C(2, 1) and radius 5 and three lines. The chord x + y = 4 cuts it at A and B; the perpendicular from C meets the chord at its midpoint M. The line y = -4 touches it at T, where CT = 5 is a radius. The line x + y = 12 misses it: the foot of the perpendicular from C is F, and CF is greater than 5
Section 01

Substitute, then solve.

A point where a line meets a circle lies on BOTH graphs, so its coordinates satisfy both equations at once. This is the same method as Topic 5 (a line and a curve): rearrange the line to make \(y\) (or \(x\)) the subject, substitute into the circle equation, expand and collect everything on one side. You get a quadratic in one variable. Solve it, then put each root back into the LINE equation to find the matching coordinate.

Worked example Two points of intersection
Find the coordinates of the points where the line \(x + y = 4\) meets the circle \(x^{2} + y^{2} - 4x - 2y - 20 = 0\).
  1. From the line, \(y = 4 - x\).
  2. Substitute: \(x^{2} + (4 - x)^{2} - 4x - 2(4 - x) - 20 = 0\).
  3. Expand: \(x^{2} + 16 - 8x + x^{2} - 4x - 8 + 2x - 20 = 0\), so \(2x^{2} - 10x - 12 = 0\).
  4. Divide by 2: \(x^{2} - 5x - 6 = 0\), so \((x - 6)(x + 1) = 0\) and \(x = 6\) or \(x = -1\).
  5. From \(y = 4 - x\): \(x = 6\) gives \(y = -2\); \(x = -1\) gives \(y = 5\). These are A and B in Fig. 1.
Answer: \((-1,\ 5)\) and \((6,\ -2)\)
Worked example A circle centred at the origin
The line \(2x - y = 5\) meets the circle \(x^{2} + y^{2} = 25\) at P and Q. Find the coordinates of P and Q.
  1. From the line, \(y = 2x - 5\) (no fractions this way round).
  2. Substitute: \(x^{2} + (2x - 5)^{2} = 25\), so \(x^{2} + 4x^{2} - 20x + 25 = 25\).
  3. \(5x^{2} - 20x = 0\), so \(5x(x - 4) = 0\) and \(x = 0\) or \(x = 4\).
  4. \(x = 0\): \(y = -5\). \(x = 4\): \(y = 3\).
  5. Check: \(0^{2} + (-5)^{2} = 25\) and \(4^{2} + 3^{2} = 25\).
Answer: \((0,\ -5)\) and \((4,\ 3)\)
Your turn
Find where the line \(y = x - 1\) meets the circle \((x - 1)^{2} + (y + 2)^{2} = 10\).
Examiner trap

Square the whole bracket: \((4 - x)^{2} = 16 - 8x + x^{2}\), not \(16 + x^{2}\) and not \(16 - x^{2}\). Then find each \(y\) from the LINE, not the circle: the circle gives two \(y\) values for each \(x\), and only one of them is on the line. Give the answer as paired coordinates.

Section 02

The discriminant decides.

The quadratic you get after substituting has one root for each meeting point, so its discriminant (from 2.3) counts them without solving. Two distinct roots: the line cuts the circle twice and is a chord (strictly, the part inside the circle is the chord). One repeated root: the line touches the circle at exactly one point and is a tangent. No real roots: the line misses the circle. Fig. 1 shows all three. Rearrange to \(ax^{2} + bx + c = 0\) FIRST, then read \(a\), \(b\) and \(c\) with their signs. Writing the equation of a tangent at a given point, and the tangents from a point outside, come in the next lesson.

\(b^{2} - 4ac\)Meeting pointsThe line is
\(> 0\)twoa chord (it cuts the circle)
\(= 0\)onea tangent (it touches the circle)
\(< 0\)noneoutside the circle (it does not meet it)
Worked example Show that a line is a tangent
Show that the line \(3x + 4y = 35\) is a tangent to the circle \((x - 2)^{2} + (y - 1)^{2} = 25\), and find the point of contact.
  1. From the line, \(y = \dfrac{35 - 3x}{4}\), so \(y - 1 = \dfrac{31 - 3x}{4}\).
  2. Substitute: \((x - 2)^{2} + \dfrac{(31 - 3x)^{2}}{16} = 25\). Multiply by 16: \(16(x^{2} - 4x + 4) + 961 - 186x + 9x^{2} = 400\).
  3. Collect: \(25x^{2} - 250x + 625 = 0\), so \(x^{2} - 10x + 25 = 0\).
  4. Discriminant: \((-10)^{2} - 4(1)(25) = 100 - 100 = 0\), one repeated root, so the line is a tangent.
  5. \((x - 5)^{2} = 0\) gives \(x = 5\), and \(y = \dfrac{35 - 15}{4} = 5\).
Answer: Discriminant \(= 0\), so the line is a tangent; the point of contact is \((5,\ 5)\)
Worked example Find k for a tangent
Find the values of \(k\) for which the line \(y = x + k\) is a tangent to the circle \(x^{2} + y^{2} = 8\). For each value, find the point of contact.
  1. Substitute: \(x^{2} + (x + k)^{2} = 8\), so \(2x^{2} + 2kx + k^{2} - 8 = 0\).
  2. Here \(a = 2,\ b = 2k,\ c = k^{2} - 8\). Tangent: \((2k)^{2} - 4(2)(k^{2} - 8) = 0\).
  3. \(4k^{2} - 8k^{2} + 64 = 0\), so \(64 - 4k^{2} = 0\), \(k^{2} = 16\) and \(k = \pm 4\).
  4. \(k = 4\): \(2x^{2} + 8x + 8 = 2(x + 2)^{2} = 0\), so \(x = -2\), \(y = 2\).
  5. \(k = -4\): \(2x^{2} - 8x + 8 = 2(x - 2)^{2} = 0\), so \(x = 2\), \(y = -2\).
Answer: \(k = 4\), touching at \((-2,\ 2)\), or \(k = -4\), touching at \((2,\ -2)\)
Your turn
Does the line \(y = 2x + 6\) meet the circle \(x^{2} + y^{2} - 2x - 4 = 0\)? Use the discriminant.
Examiner trap

Use the discriminant of the quadratic you get AFTER substituting and collecting every term on one side, not of the circle equation or the line. In the \(k\) question, \(c\) is the whole expression \(k^{2} - 8\), and \((2k)^{2} = 4k^{2}\), not \(2k^{2}\). A tangent has TWO answers for \(k\) here: \(k^{2} = 16\) means \(k = 4\) or \(k = -4\).

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Substituting \(y = x + 1\) into \(x^{2} + y^{2} = 13\) and simplifying gives

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A line is substituted into the equation of a circle and the resulting quadratic has discriminant \(0\). The line is

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A circle has centre \((3,\ 4)\) and radius 5. The foot of the perpendicular from the centre to a line L is \((6,\ 8)\). What is L?

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M is the midpoint of a chord AB of a circle with centre C. Which statement is always true?

Section 03

The perpendicular from the centre.

The perpendicular from the centre of a circle to a chord bisects the chord, so it gives the chord's midpoint, and it also measures how far the line is from the centre. To find that perpendicular distance, write the equation of the line through the centre C perpendicular to the given line (gradients multiply to \(-1\)), solve the two line equations together to get the foot F, then find CF with the distance formula. If \(CF < r\) the line is a chord, if \(CF = r\) it is a tangent, if \(CF > r\) it misses. For a chord, F is the midpoint M and Pythagoras gives half the chord: \(\left(\tfrac{1}{2}AB\right)^{2} = r^{2} - CM^{2}\). Compare squares to avoid surds.

Worked example The distance test
Without solving a quadratic, show that the line \(x + y = 12\) does not meet the circle \((x - 2)^{2} + (y - 1)^{2} = 25\).
  1. The line \(y = 12 - x\) has gradient \(-1\), so the perpendicular through C(2, 1) has gradient \(1\): \(y - 1 = x - 2\), that is \(y = x - 1\).
  2. Foot F: \(x + (x - 1) = 12\), so \(2x = 13\), \(x = 6.5\) and \(y = 5.5\).
  3. \(CF^{2} = (6.5 - 2)^{2} + (5.5 - 1)^{2} = 4.5^{2} + 4.5^{2} = 40.5\).
  4. \(r^{2} = 25\) and \(40.5 > 25\), so \(CF > r\): the line is outside the circle (Fig. 1).
Answer: \(CF^{2} = 40.5 > 25 = r^{2}\), so the line does not meet the circle
Worked example Midpoint and length of a chord
The line \(x + y = 4\) is a chord of the circle \((x - 2)^{2} + (y - 1)^{2} = 25\). Find the midpoint M of the chord and the length of the chord.
  1. The perpendicular through C(2, 1) is again \(y = x - 1\).
  2. M: \(x + (x - 1) = 4\), so \(x = 2.5\) and \(y = 1.5\).
  3. \(CM^{2} = 0.5^{2} + 0.5^{2} = 0.5\), so \(\left(\tfrac{1}{2}AB\right)^{2} = 25 - 0.5 = 24.5 = \dfrac{49}{2}\).
  4. \(\tfrac{1}{2}AB = \dfrac{7}{\sqrt{2}}\), so \(AB = \dfrac{14}{\sqrt{2}} = 7\sqrt{2}\).
  5. Check with A(-1, 5) and B(6, -2) from Section 1: midpoint \((2.5,\ 1.5)\), \(AB = \sqrt{7^{2} + 7^{2}} = 7\sqrt{2}\).
Answer: \(M(2.5,\ 1.5)\), \(AB = 7\sqrt{2}\)
Your turn
The line \(y = 2x - 5\) cuts \(x^{2} + y^{2} = 25\) at P and Q. Use the perpendicular from O to find the midpoint of PQ and the length PQ.
Examiner trap

The distance from the centre to the line is the PERPENDICULAR distance CF. The distance from C to some other point on the line, such as where it crosses an axis, is longer and proves nothing. Compare like with like: \(CF^{2}\) with \(r^{2}\), or \(CF\) with \(r\), never \(CF^{2}\) with \(r\). And \(r^{2} - CM^{2}\) gives HALF the chord squared: double the square root at the end.

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Find the coordinates of the points of intersection of the line \(x - 2y + 1 = 0\) and the circle \(x^{2} + y^{2} - 6x - 4y + 8 = 0\).

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Find the values of \(k\) for which the line \(y = 2x + k\) is a tangent to the circle \(x^{2} + y^{2} = 5\).

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Find the set of values of \(k\) for which the line \(y = x + k\) does not meet the circle \(x^{2} + y^{2} = 18\).

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The line \(x + 2y = 5\) meets the circle \(x^{2} + y^{2} = 20\) at A and B. Find the length of AB.

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Six original Cambridge-style questions
Classifying a line, points of intersection with the midpoint and length of the chord, showing a line is a tangent, values of \(k\) for a tangent and a chord, the perpendicular distance test and a pair of tangents from one point.