On, inside or outside
x² + y² = 10
(1, 3): 1 + 9 = 10, on it
(2, 2): 4 + 4 = 8, less than 10: inside
(3, 2): 9 + 4 = 13, more than 10: outside
Higher · AQA · 45 minutes
Students grow a circle until it just touches a line, then build x² + y² = r² from Pythagoras and find tangents as lines at right angles to the radius. Slides to present, a vote that shows who's stuck, practice and homework.
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x² + y² = k is a circle (k is r², the radius squared). Move k: which circle just touches the line x + 2y = 5, and where? What do you notice about the radius to that point?
They don't cross on these axes
A point (x, y) is a distance √(x² + y²) from (0, 0), by Pythagoras. So the points exactly r from (0, 0), a circle, are those with x² + y² = r². A point with x² + y² less than r² is inside; more than r², outside. The tangent at a point on the circle is at right angles to the radius there: its gradient is the radius's gradient flipped, with its sign changed.
The lesson goes picture first. Students grow a circle until it touches a line before anyone explains, and notice the radius meets the line at a right angle. Then they build the equation from Pythagoras, test points against it, and find tangents with the perpendicular gradients of the lesson before. It is Higher; the stretch finds where a line crosses a circle.
Students should use Pythagoras' theorem, find a perpendicular gradient and the line through a point with a given gradient (the lesson Parallel and perpendicular lines).
Taught in Parallel and perpendicular lines.
60 minutes in total. Each step says what changes for this length.
5 min 1 · Hook · students think
Students predict which circle just touches a line, and where, before anyone explains.
“x² + y² = k is a circle around (0, 0). The line x + 2y = 5 stays put. Which k makes the circle just touch the line, and where? Write a guess.”
Guessing k = 5 only because of the 5 in x + 2y = 5: it is right here, but a guess is not a reason. Ask for the point where it touches, and why there.
Give a minute to guess, then move k from 2 upwards. At k = 5 the circle just touches the line at (1, 2); past 5 it cuts it twice.
Don't explain yet. Ask about the radius to (1, 2): its gradient is 2, the line's is −½. They met that pair in the last lesson.
Gather guesses for k with Ask the class, then move k on the board to 5. Keep the point (1, 2) for the explaining step.
Shorter: Move k to 5 straight away; skip the guesses (−2 min).
12 min 2 · Explain · you explain
Build x² + y² = r² from Pythagoras, test points, then the tangent at right angles to the radius.
“Every point on a circle is the same distance from its centre. Pythagoras turns that distance into an equation.”
Reading r² as the radius: x² + y² = 16 has radius 4.
On, inside or outside
x² + y² = 10
(1, 3): 1 + 9 = 10, on it
(2, 2): 4 + 4 = 8, less than 10: inside
(3, 2): 9 + 4 = 13, more than 10: outside
The tangent at (1, 3)
x² + y² = 10, the point (1, 3)
Radius gradient: 3 ÷ 1 = 3
Tangent gradient: −⅓
3 = −⅓ × 1 + c, so c = 10/3
y = −⅓x + 10/3, or x + 3y = 10
First slide: a point (x, y) is x across and y up from (0, 0), so its distance is √(x² + y²). (3, 4) is √25 = 5 away, and so are (4, 3), (5, 0) and (0, 5): all on x² + y² = 25. Every point 5 away is on it, and only those: a circle.
Second slide: x² + y² = r². Move r: the number on the right is r², so x² + y² = 16 has radius 4 and x² + y² = 10 has radius √10.
Third slide: to test a point, work out x² + y² and compare it with r². Equal: on. Less: inside. More: outside. Never compare the coordinates with r.
Fourth slide: the tangent at (1, 3). The radius lies along y = 3x, gradient 3. The tangent is at right angles to the radius (a circle theorem), so its gradient is −⅓. Put (1, 3) in to find c. This answers the hook: the line x + 2y = 5 touches x² + y² = 5 at (1, 2), where the radius has gradient 2 and the line −½.
Vote before moving r (Ask the class): r to 4, which equation? Then the three points: on, inside or outside?
Shorter: Skip the votes; one point tested aloud (−4 min).
8 min 3 · Your turn · students write
Each worked example beside a twin that students solve themselves.
“Watch me do one. Then you do its twin, which is almost the same.”
Squaring a negative coordinate as a negative: (−5)² = 25.
Inside or outside
(4, −2) and x² + y² = 25
16 + 4 = 20, less than 25: inside
Then theirs
(−5, 1) and x² + y² = 25
The circle through a point
Centre (0, 0), through (2, −4)
4 + 16 = 20, so x² + y² = 20
Then theirs
Centre (0, 0), through (−3, 1)
A tangent
x² + y² = 25 at (3, 4)
Radius gradient 4/3, so the tangent's is −¾
4 = −¾ × 3 + c, so c = 25/4
y = −¾x + 25/4, or 3x + 4y = 25
Then theirs
x² + y² = 13 at (2, 3)
Work the first example aloud, writing each line. Then students solve its twin while you watch who squares −5 as −25.
Do the same for each pair. The third pair catches students who give the tangent the radius's gradient.
Open the answers slide once most have finished their twins.
Shorter: Do the first and third pairs only (−3 min).
25 min 4 · Who's stuck? · students vote
One question shows who's stuck; then practice in levels, and a stretch.
For this length: The stretch, then: how many times can a line meet a circle?
“Choose an answer on your own. Nobody talks until everyone has chosen.”
Giving the tangent the radius's gradient.
Show your working on paper. Write fractions with /, such as -1/2, and negative numbers with a minus sign. The stretch is optional.
The question that shows who's stuck
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
Support · For question 3: divide every term by 2 first, so it reads x² + y² = … . The number on the right is r²: take its square root for the radius.
Everyone chooses before anyone explains. If they chose the radius's own gradient, ask which line has it. If they changed only the sign, or only flipped, ask them to multiply their answer by the radius's gradient: is it −1?
Questions 1–2 build fluency; 3–4 change the look; 5–6 ask for reasons; 7–8 are tangents, as exam questions ask them; 9 is a tangent where the radius has no gradient to flip. The stretch finds where a line crosses a circle.
Students show their working on paper or the board; the quiz keeps their final answers.
If the class is split, sketch the radius to (2, 4), then a line through (2, 4) with each gradient chosen: which one crosses the radius at a right angle? Then practise.
Shorter: After the vote, questions 1–4 only (−6 min).
Longer: The stretch, then: how many times can a line meet a circle? (+10 min).
10 min 5 · Exit · students show
One tangent and one sentence show who links the tangent's gradient to the radius.
For this length: Check it: put the tangent into the circle; it meets it once
“Find the tangent to x² + y² = 17 at (1, 4). Then say in one sentence why you used that gradient.”
A tangent through (0, 0): it goes through (1, 4).
Show your working; one sentence for the second part.
Find the equation of the tangent to x² + y² = 17 at the point (1, 4).
Then say why the tangent has the gradient you used.
The sentences show who knows why the gradients are linked: start the next lesson with the hook's circle again if few mention the right angle. Set the homework.
Longer: Check it: put the tangent into the circle; it meets it once (+5 min).
One to one: let the student grow the circle and say where it touches before you explain; for each tangent, ask them to say the radius's gradient aloud before they flip it.
Show your working. For each point, work out x² + y² and compare it with r². For each tangent, write the radius's gradient first.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
A point (x, y) is x across and y up from (0, 0), so by Pythagoras its distance from (0, 0) is √(x² + y²). A circle is every point the same distance r from its centre, so its points are exactly those with x² + y² = r². Testing (3, 4) on x² + y² = 25 makes it concrete.
It is a circle theorem: the tangent at a point is perpendicular to the radius there. One way to see it: the point where the tangent touches is the point of the line closest to the centre, and the shortest way from a point to a line meets it at a right angle. The hook shows it as a growing circle that first touches a line.
Find the gradient of the radius from (0, 0) to the point: y ÷ x. Flip it and change its sign for the tangent's gradient. Put the point into y = mx + c to find c. Students lose marks by using the radius's gradient, or by putting (0, 0) in instead of the point.
Higher. GCSE asks for circles centred at the origin, x² + y² = r², and tangents at a point; circles with other centres come at A-level and in Level 2 Further Maths. The lesson needs perpendicular gradients, from Parallel and perpendicular lines.
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Curriculum reference: AQA GCSE Mathematics 8300: A16 (the equation of a circle with centre at the origin, and the equation of a tangent to a circle at a given point, Higher). It uses A9 (perpendicular gradients).
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