A circle has centre (0, 0) and passes through the point (0, -7). Decide whether the point (5, 5) lies inside, on or outside this circle.

Outside, because 5^2 + 5^2 = 50 and 50 > 49

Inside, because both coordinates are less than the radius 7

On the circle, because 5^2 + 5^2 = 50 and the radius is sqrt(50)

Inside, because 5 + 5 = 10 and 10 < 49

A drone flies in a straight line along the path y = x - 3 across a grid in which one unit represents 1 km. A circular no-fly zone has centre the origin and boundary x^2 + y^2 = 29. Find the coordinates of the two points where the drone's path crosses the boundary. Give the point with the larger x-coordinate first.

x of first point = 5, y of first point = 2, x of second point = -2, y of second point = -5

Is the equation of a circle on the Foundation paper?

No. Spec point A16 is bold in the 1MA1 specification, which marks it as Higher tier only, so both the equation x^2 + y^2 = r^2 and the tangent to a circle at a point are Higher content. Foundation candidates do meet circle theorems for angles at G10 in the crossover topics, but never the coordinate version.

Is the equation of a circle on the Edexcel formulae sheet?

No. The Exam Aid sheet issued with each 1MA1 paper gives the circumference and area of a circle, the quadratic formula, Pythagoras' theorem and the trigonometric ratios, the sine and cosine rules, the area of a triangle, compound interest and two probability rules. Neither x^2 + y^2 = r^2 nor the perpendicular gradient rule appears on it, so both are recall.

How do you find the equation of a tangent to a circle?

Work out the gradient of the radius from the origin to the point of contact, which is just the y-coordinate divided by the x-coordinate. Take the negative reciprocal for the gradient of the tangent, since a tangent meets a radius at 90^ . Then substitute the point of contact into y = mx + c to find c. For (8, 6) on x^2 + y^2 = 100 that gives m = -(4)/(3) and c = (50)/(3).

How do you show that a line is a tangent to a circle?

Substitute the line into the circle equation and solve the quadratic that comes out. One repeated root means the line touches at exactly one point, so it is a tangent; two roots means it cuts the circle twice and no real roots means it misses. For example y = x + 8 in x^2 + y^2 = 32 gives 2x^2 + 16x + 32 = 0, which factorises to 2(x + 4)^2 = 0, a repeated root at x = -4.

Equation of a Circle - Edexcel GCSE Higher Maths

Typically 4-6 marks when it appears, usually one tangent question late in the paper · Spec 1MA1

A circle with centre the origin and radius $r$ has equation $x^2 + y^2 = r^2$ . That equation, plus finding the tangent to such a circle at a given point, is the whole of spec point A16 on Edexcel's 1MA1 specification, and A16 is printed in bold, which means Higher tier only. The equation is Pythagoras' theorem wearing different clothes: $x$ and $y$ are the two short sides of a right-angled triangle whose hypotenuse is the radius, so a point sits on the circle exactly when the square of its distance from the origin comes to $r^2$ . GCSE never moves the centre off $(0, 0)$ , so if you meet $(x - 3)^2 + (y + 1)^2 = 16$ you are reading an A level book.

Edexcel sets this at two very different sizes. The small version is a single mark near the middle of a paper: write down the radius of $x^2 + y^2 = 36$ , or write down the equation of the circle through $(0, 7)$ with centre the origin. The large version sits in the last third of the paper and runs to 4-5 marks: the tangent at a stated point, then something extra, such as where that tangent crosses an axis or the area of the triangle it makes with the origin. Because A16 is Higher only it can appear on any of the three papers, and Paper 1 being non-calculator matters here, since $x^2 + y^2 = 40$ has radius $\sqrt{40}$ and a tangent gradient is far more often a fraction such as $-\frac{3}{4}$ than a whole number. Where a line meets a circle in two places you are solving one linear and one quadratic equation, which is the method set out on the simultaneous equations page.

Three things account for most of the lost marks. Reading the radius of $x^2 + y^2 = 45$ as 45 rather than $\sqrt{45}$ costs the first mark and everything built on it. Working out the gradient of the radius and then writing that number into the tangent equation skips the perpendicular step entirely, and the examiner sees a line that cuts straight through the circle. Third, the constant: the tangent is $y = mx + c$ and $c$ has to come from substituting the point of contact, not from the circle. Practise the four-line routine until it is automatic: gradient of the radius, negative reciprocal, substitute the point, state the equation.

Worked example

The circle

C

has equation

x^2 + y^2 = 100

. The point

A(6, -8)

lies on

C

. The tangent to

C

A

crosses the

y

-axis at the point

B

. Work out the coordinates of

B

[5 marks]

Check $A$ is on the circle and find the gradient of the radius $OA$ : $6^2 + (-8)^2 = 100$ , and the gradient is $\frac{-8 - 0}{6 - 0} = -\frac{4}{3}$ .
M1: gradient of the radius from $(0, 0)$ to the given point, change in $y$ over change in $x$
The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of $-\frac{4}{3}$ , which is $\frac{3}{4}$ .
M1: using $m_1 m_2 = -1$ ; invert the fraction and change the sign, both in one move
Substitute $A(6, -8)$ into $y = \frac{3}{4}x + c$ : $-8 = \frac{3}{4} \times 6 + c$ , so $-8 = 4.5 + c$ and $c = -12.5$ .
M1: substituting the point of contact into $y = mx + c$ with their tangent gradient
The tangent is $y = \frac{3}{4}x - 12.5$ .
A1: a correct tangent equation; a negative $c$ is expected here because $A$ is below the $x$ -axis
The $y$ -axis is $x = 0$ , so $y = -12.5$ and $B$ is the point $(0, -12.5)$ .
A1: coordinates, not just the number; $c$ is the $y$ -intercept, so no extra working was needed

Practice questions

These are original questions in Edexcel 1MA1 Higher style. There are no diagrams, so each point is named with its coordinates and each circle with its equation. Tangent questions ask for $m$ and $c$ as two separate values, which is what the mark scheme wants; type fractions directly if you prefer them to decimals (-3/4 and -0.75 both mark correct). The line-and-circle questions here are the same linear-plus-quadratic work as the simultaneous equations set, so do those alongside these.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [2 marks]

$P$ is a point on a circle with centre the origin $O$ . The radius $OP$ has gradient $\frac{2}{3}$ . The tangent to the circle at $P$ is perpendicular to $OP$ . Work out the gradient of the tangent.

Your answer (fractions like 1/2 are fine)
Question 2 Warm-up [2 marks]

The circle $x^2 + y^2 = 20$ has radius $r$ . Work out the value of $r$ , giving your answer correct to 2 decimal places.

Your answer (fractions like 1/2 are fine)
Question 3 Exam pace [3 marks]

The point $P(a, 6)$ , where $a > 0$ , lies on a circle with centre the origin. The tangent to the circle at $P$ has gradient $-\frac{2}{3}$ . Find the value of $a$ .

Your answer (fractions like 1/2 are fine)
Question 4 Exam pace [4 marks]

The point $(4, 2)$ lies on the circle $x^2 + y^2 = 20$ . Find the equation of the tangent to the circle at $(4, 2)$ in the form $y = mx + c$ . Give the values of $m$ and $c$ .

m = c =
Question 5 Exam pace [3 marks]

A circle has centre $(0, 0)$ and passes through the point $(0, -7)$ . Decide whether the point $(5, 5)$ lies inside, on or outside this circle.

Answer options for question 5
Outside, because $5^2 + 5^2 = 50$ and $50 > 49$ Inside, because both coordinates are less than the radius 7 On the circle, because $5^2 + 5^2 = 50$ and the radius is $\sqrt{50}$ Inside, because $5 + 5 = 10$ and $10 < 49$
Question 6 Stretch [5 marks]

A drone flies in a straight line along the path $y = x - 3$ across a grid in which one unit represents 1 km. A circular no-fly zone has centre the origin and boundary $x^2 + y^2 = 29$ . Find the coordinates of the two points where the drone's path crosses the boundary. Give the point with the larger $x$ -coordinate first.

x of first point = y of first point = x of second point = y of second point =

Common mistakes examiners see

Taking the number on the right of the equation as the radius, so $x^2 + y^2 = 64$ is given radius 64 instead of 8, and $x^2 + y^2 = 45$ is given 45 instead of $\sqrt{45}$ .

Read the equation as $x^2 + y^2 = r^2$ every time and square root the right-hand side. When the number is not a square, leave the answer as a surd on Paper 1; $\sqrt{45}$ is exact and 6.7 is not.
Writing the radius equation the wrong way round when the radius is given: answering $x^2 + y^2 = 6$ for a circle of radius 6.

Square the radius before it goes into the equation, so radius 6 gives $x^2 + y^2 = 36$ . Test your answer by putting the point $(6, 0)$ in; it must satisfy the equation, because it is on the circle.
Using the gradient of the radius as the gradient of the tangent. At $(3, 4)$ on $x^2 + y^2 = 25$ that produces $y = \frac{4}{3}x$ , a line through the centre rather than a tangent.

Write both gradients down on separate lines, labelled: radius $\frac{4}{3}$ , tangent $-\frac{3}{4}$ . If your tangent goes through $(0, 0)$ you have found the radius, since a tangent to a circle centred on the origin never passes through the origin.
Leaving the tangent as $y = mx$ because the circle is centred on the origin, or reusing $r^2$ as the intercept.

Find $c$ by substituting the point of contact into $y = mx + c$ . At $(6, 8)$ with $m = -\frac{3}{4}$ that is $8 = -4.5 + c$ , so $c = 12.5$ , and it is that substitution the third method mark is for.
Comparing with the radius instead of its square when deciding whether a point is inside or outside. For $(7, -2)$ and $x^2 + y^2 = 50$ , saying 'outside, because $7 + 2 > \sqrt{50}$ ' rather than working out $7^2 + (-2)^2 = 53$ .

Work out $x^2 + y^2$ for the point and compare that single number with the $r^2$ in the equation: smaller means inside, equal means on, larger means outside. Squaring a negative coordinate makes it positive, so $(-6, 8)$ and $(6, 8)$ give the same 100.

Frequently asked questions

Is the equation of a circle on the Foundation paper?: No. Spec point A16 is bold in the 1MA1 specification, which marks it as Higher tier only, so both the equation $x^2 + y^2 = r^2$ and the tangent to a circle at a point are Higher content. Foundation candidates do meet circle theorems for angles at G10 in the crossover topics, but never the coordinate version.
Is the equation of a circle on the Edexcel formulae sheet?: No. The Exam Aid sheet issued with each 1MA1 paper gives the circumference and area of a circle, the quadratic formula, Pythagoras' theorem and the trigonometric ratios, the sine and cosine rules, the area of a triangle, compound interest and two probability rules. Neither $x^2 + y^2 = r^2$ nor the perpendicular gradient rule appears on it, so both are recall.
How do you find the equation of a tangent to a circle?: Work out the gradient of the radius from the origin to the point of contact, which is just the $y$ -coordinate divided by the $x$ -coordinate. Take the negative reciprocal for the gradient of the tangent, since a tangent meets a radius at $90^\circ$ . Then substitute the point of contact into $y = mx + c$ to find $c$ . For $(8, 6)$ on $x^2 + y^2 = 100$ that gives $m = -\frac{4}{3}$ and $c = \frac{50}{3}$ .
How do you show that a line is a tangent to a circle?: Substitute the line into the circle equation and solve the quadratic that comes out. One repeated root means the line touches at exactly one point, so it is a tangent; two roots means it cuts the circle twice and no real roots means it misses. For example $y = x + 8$ in $x^2 + y^2 = 32$ gives $2x^2 + 16x + 32 = 0$ , which factorises to $2(x + 4)^2 = 0$ , a repeated root at $x = -4$ .

Equation of a Circle - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics