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P1 3

Coordinate geometry

Pure Mathematics 1 · Paper 1 · AS and A Level

🎯What you need to be able to do

  • Find the length, gradient and midpoint of a line segment.
  • Write the equation of a straight line in any of the standard forms.
  • Use the parallel and perpendicular gradient conditions, including for perpendicular bisectors.
  • Find where two lines, or a line and a curve, intersect.
  • Write the equation of a circle in both centre–radius and expanded form, and convert between them.
  • Use the circle properties: the tangent is perpendicular to the radius, and the perpendicular from the centre bisects a chord.
  • Solve problems about the intersection of a line and a circle, including tangency.
  • Transform a relationship into a straight line to find unknown constants.

📚The mathematics

The basics of a line segment

\( \text{length} = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}} \)
\( m = \dfrac{y_2-y_1}{x_2-x_1} \)
\( \text{midpoint} = \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right) \)

The gradient is the one people get upside down under pressure. It is the change in \(y\) over the change in \(x\), taken in the same order for both — subtracting the points in opposite orders top and bottom flips the sign.

Equations of straight lines

\( y = mx + c \)
\( y - y_1 = m(x - x_1) \)
\( ax + by + c = 0 \)

Use point–gradient form to build a line, because it needs only a point and a gradient and no rearranging. Cambridge often asks for the answer in the form \( ax + by + c = 0 \) with integer coefficients, so read the instruction and clear the fractions if it does.

Parallel: \( m_1 = m_2 \). Perpendicular: \( m_1 m_2 = -1 \), so the perpendicular gradient is the negative reciprocal — flip it and change the sign. A gradient of \( \tfrac{2}{5} \) gives \( -\tfrac{5}{2} \). Doing only half of that (flipping without the sign change, usually) is the standard error.

The perpendicular bisector of \(AB\) combines both ideas: it passes through the midpoint of \(AB\) and has the negative reciprocal of the gradient of \(AB\). It is the set of points equidistant from \(A\) and \(B\), which is why it appears in circle questions — the centre of a circle through two points lies on their perpendicular bisector.

The circle

\[ (x - a)^{2} + (y - b)^{2} = r^{2} \]

Centre \( (a, b) \), radius \(r\). Watch the signs: \( (x-3)^{2} + (y+2)^{2} = 25 \) has centre \( (3, -2) \) — the opposite sign to the one written — and radius 5, not 25.

The expanded form \( x^{2} + y^{2} + 2gx + 2fy + c = 0 \) is converted back by completing the square in \(x\) and in \(y\) separately:

\[ x^{2} + y^{2} - 6x + 4y - 12 = 0 \;\Longrightarrow\; (x-3)^{2} - 9 + (y+2)^{2} - 4 - 12 = 0 \;\Longrightarrow\; (x-3)^{2} + (y+2)^{2} = 25 \]

Two geometric properties carry most circle questions, and both turn a circle problem into a straight-line problem:

  • A tangent is perpendicular to the radius at the point of contact. So to find a tangent, find the gradient of the radius to that point and take the negative reciprocal. You almost never need calculus for a circle tangent at P1.
  • The perpendicular from the centre to a chord bisects it. That creates a right-angled triangle with the radius as hypotenuse, half the chord as one leg, and the distance from centre to chord as the other — so Pythagoras finishes the job.

Line meets circle

Substitute the line into the circle and you get a quadratic, and its discriminant tells you the geometry — exactly as in P1 1:

\( b^{2}-4ac > 0 \) — the line is a chord
\( b^{2}-4ac = 0 \) — the line is a tangent
\( b^{2}-4ac < 0 \) — the line misses

An alternative for tangency that is often quicker: the line is a tangent exactly when the perpendicular distance from the centre to the line equals the radius.

Transforming a relationship to a straight line

A relationship that is not linear can often be made linear by choosing the right variables, and then the gradient and intercept give you the unknown constants.

  • \( y = ax^{n} \): take logarithms, \( \lg y = \lg a + n \lg x \). Plot \( \lg y \) against \( \lg x \) — gradient \(n\), intercept \( \lg a \).
  • \( y = ab^{x} \): \( \lg y = \lg a + x\lg b \). Plot \( \lg y \) against \(x\) — gradient \( \lg b \), intercept \( \lg a \).

The intercept gives you \( \lg a \), not \(a\): you must undo the logarithm at the end. Forgetting that final step is the commonest slip on this sub-topic.

✏️Worked example

A circle has equation \( x^{2} + y^{2} - 8x + 2y - 8 = 0 \). (a) Find its centre and radius. (b) Show that the point \( P(7, 3) \) lies on the circle, and find the equation of the tangent at \(P\), giving your answer in the form \( ax + by + c = 0 \). (c) The line \( y = 2x + k \) is a tangent to the circle. Find the possible values of \(k\).

(a) Complete the square in each variable:

\[ (x-4)^{2} - 16 + (y+1)^{2} - 1 - 8 = 0 \;\Longrightarrow\; (x-4)^{2} + (y+1)^{2} = 25 \]

So the centre is \( C(4, -1) \) and the radius is 5.

(b) Substitute \(P\) into the left-hand side: \( 49 + 9 - 56 + 6 - 8 = 0 \). ✓ So \(P\) lies on the circle.

The radius \(CP\) has gradient \( \dfrac{3 - (-1)}{7 - 4} = \dfrac{4}{3} \). The tangent is perpendicular to it, so its gradient is \( -\dfrac{3}{4} \). Using point–gradient form at \(P\):

\[ y - 3 = -\tfrac{3}{4}(x - 7) \;\Longrightarrow\; 4y - 12 = -3x + 21 \;\Longrightarrow\; 3x + 4y - 33 = 0 \]

(c) Substitute \( y = 2x + k \) into the circle:

\[ (x-4)^{2} + (2x + k + 1)^{2} = 25 \]

Expanding, \( x^{2} - 8x + 16 + 4x^{2} + 4x(k+1) + (k+1)^{2} = 25 \), which collects to

\[ 5x^{2} + \left(4k - 4\right)x + \left[(k+1)^{2} - 9\right] = 0 \]

Tangency requires the discriminant to be zero:

\[ (4k-4)^{2} - 20\left[(k+1)^{2} - 9\right] = 0 \]

So \( 16k^{2} - 32k + 16 - 20k^{2} - 40k - 20 + 180 = 0 \), giving \( -4k^{2} - 72k + 176 = 0 \), and dividing by \( -4 \), \( k^{2} + 18k - 44 = 0 \). Hence \( k = \dfrac{-18 \pm \sqrt{324 + 176}}{2} = \dfrac{-18 \pm \sqrt{500}}{2} = -9 \pm 5\sqrt{5} \).

Check it with the other method. The perpendicular distance from \( (4,-1) \) to \( 2x - y + k = 0 \) must equal 5: \( \dfrac{|2(4) - (-1) + k|}{\sqrt{5}} = 5 \), so \( |9 + k| = 5\sqrt{5} \) and \( k = -9 \pm 5\sqrt{5} \). Same answer, far less algebra — worth knowing as a check even when the question steers you to the discriminant.
Radius versus radius squared, and the sign of the centre. In (a), reading the radius as 25 rather than 5 wrecks every later part. And a circle written \( (x-4)^{2} + (y+1)^{2} \) has centre \( (4, -1) \): the \(y\)-coordinate flips sign but the \(x\)-coordinate does not, because of the \( + \) in the bracket. Write the centre down immediately after completing the square, and check it against a rough sketch.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. \( A(-1, 4) \) and \( B(5, -2) \). Find the equation of the perpendicular bisector of \(AB\) in the form \( ax + by + c = 0 \).
Midpoint: \( \left(\tfrac{-1+5}{2}, \tfrac{4-2}{2}\right) = (2, 1) \). Gradient of \(AB\): \( \dfrac{-2-4}{5-(-1)} = \dfrac{-6}{6} = -1 \). Perpendicular gradient: \( +1 \). So \( y - 1 = 1(x - 2) \), giving \( y = x - 1 \), or \( x - y - 1 = 0 \).
2. Find the centre and radius of the circle \( x^{2} + y^{2} + 10x - 4y + 13 = 0 \).
Complete the square separately: \( (x+5)^{2} - 25 + (y-2)^{2} - 4 + 13 = 0 \), so \( (x+5)^{2} + (y-2)^{2} = 16 \). Centre \( (-5, 2) \), radius 4. Note both signs flip relative to the brackets, and the constant 13 moves across with the two subtracted squares.
3. A circle has centre \( (2, 5) \) and radius 10. A chord of the circle has length 16. Find the perpendicular distance from the centre to the chord.
The perpendicular from the centre bisects the chord, giving a right-angled triangle with hypotenuse the radius 10 and one leg half the chord, 8. So the distance \(d\) satisfies \( d^{2} + 8^{2} = 10^{2} \), giving \( d^{2} = 36 \) and \( d = 6 \). The centre coordinates are not needed at all — a reminder to check what the question actually requires before using everything it gives you.
4. The line \( y = 3x - 1 \) meets the curve \( y = x^{2} + x + 2 \). Find the coordinates of the points of intersection.
Equate: \( x^{2} + x + 2 = 3x - 1 \), so \( x^{2} - 2x + 3 = 0 \). The discriminant is \( 4 - 12 = -8 < 0 \), so there are no real solutions — the line does not meet the curve at all. Stating that, with the discriminant as the reason, is the complete answer. Manufacturing coordinates here would score nothing.
5. The point \( (6, 8) \) lies on a circle with centre \( (2, 5) \). Find the equation of the tangent at that point.
Gradient of the radius: \( \dfrac{8-5}{6-2} = \dfrac{3}{4} \). The tangent is perpendicular, so its gradient is \( -\dfrac{4}{3} \). Then \( y - 8 = -\tfrac{4}{3}(x - 6) \), giving \( 3y - 24 = -4x + 24 \), so \( 4x + 3y - 48 = 0 \). (As a check, the radius is \( \sqrt{16+9} = 5 \), a 3–4–5 triangle, which is why the numbers are tidy.)
6. Variables \(x\) and \(y\) are related by \( y = ax^{n} \). When \( \lg y \) is plotted against \( \lg x \) a straight line of gradient 1.5 passing through \( (0, 0.7) \) is obtained. Find \(a\) and \(n\).
Taking logarithms: \( \lg y = \lg a + n\lg x \), which is a straight line with gradient \(n\) and intercept \( \lg a \). So \( n = 1.5 \), and \( \lg a = 0.7 \), giving \( a = 10^{0.7} = 5.01 \). The last step is the one candidates omit: the intercept is \( \lg a \), not \(a\), so it must be undone with a power of 10.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — plot a circle and a line with a slider on the intercept, and watch tangency happen
  • GeoGebra — drag a chord and see the perpendicular from the centre bisect it
  • Khan Academy — equations of circles