The shape of the lesson
By the end of the lesson, learners can
- Centre and radius: A point \((x, y)\) is on the circle when its distance from the centre \((a, b)\) is \(r\).
- The expanded form: Expanding the brackets gives \(x^{2} + y^{2} + 2gx + 2fy + c = 0\).
- Tangents and circle geometry: In the lab, move the point of contact around the circle and check that the tangent always meets the radius at \(90^{\circ}\).
- Centre and radius give the equation directly; remember \(r^{2}\) on the right.
- Expanded form: make the \(x^{2}\) and \(y^{2}\) coefficients 1, then complete both squares.
- Diameter: centre = midpoint, radius = half the length.
- Tangent at \(P\): perpendicular to the radius \(CP\).
- Inside or outside: compare the distance from the centre with \(r\).
One picture every idea returns to
\((x - a)^{2} + (y - b)^{2} = r^{2}\) is the circle with centre \((a, b)\) and radius \(r\). For \(x^{2} + y^{2} + 2gx + 2fy + c = 0\), complete the square: the centre is \((-g, -f)\) and the radius is \(\sqrt{g^{2} + f^{2} - c}\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Pythagoras, all the way round.. Every point on a circle is the same distance from its centre, and Pythagoras turns that fact into an equation. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Centre and radius | A point \((x, y)\) is on the circle when its distance from the centre \((a, b)\) is \(r\). Model the worked example Write the equation on the board, one line at a time. Your turn: Find the equation of the circle with centre \((2, -1)\) that passes through \((5, 3)\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: The expanded form | Expanding the brackets gives \(x^{2} + y^{2} + 2gx + 2fy + c = 0\). Model the worked example Complete two squares on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Tangents and circle geometry | In the lab, move the point of contact around the circle and check that the tangent always meets the radius at \(90^{\circ}\). Project tab 3 of the lab (Circle theorems lab, tangents) and let a student drive it while the class predicts. Model the worked example Tangent at a point on the board, one line at a time. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 22 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Tangents and circle geometry, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The right-hand side is \(r^{2}\), not \(r\): a circle of radius 5 has \(= 25\), and \((x - 3)^{2} + (y + 2)^{2} = 12\) has radius \(\sqrt{12} = 2\sqrt{3}\), not 12. | Return to Centre and radius: ask a pair to state the correct version and show it on the lab before moving on. |
| The coefficients of \(x^{2}\) and \(y^{2}\) must be 1 before you read off \(g\) and \(f\). | For \(2x^{2} + 2y^{2} - 8x + 12y - 6 = 0\), divide by 2 first. And watch the signs: \(-6x\) gives centre \(x = +3\). |
| The tangent's gradient is the negative reciprocal of the RADIUS gradient, not of the gradient from the origin to \(P\). | Always use the centre, which is only the origin for \(x^{2} + y^{2} = r^{2}\). |
The equation of a circle: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Write down the equation of the circle with centre \((-3, 1)\) and radius \(\sqrt{7}\), and express it in the form \(x^{2} + y^{2} + 2gx + 2fy + c = 0\).
Find the centre and radius of the circle \(2x^{2} + 2y^{2} - 8x + 12y - 6 = 0\).
The circle \(x^{2} + y^{2} - 4x - 2y - 20 = 0\) passes through \(P(-1, 5)\). Find the equation of the tangent to the circle at \(P\).
The points \(A(-1, 1)\), \(B(9, 1)\) and \(C(1, 5)\) are given.
(a) Show that angle \(ACB = 90^{\circ}\).
(b) Hence find the equation of the circle through \(A\), \(B\) and \(C\).
(c) Find the points where this circle crosses the \(y\)-axis.
Show that the point \((7, 3)\) lies outside the circle \(x^{2} + y^{2} - 6x + 2y - 15 = 0\).
Coordinates are in metres. A sprinkler at \(S(4, 3)\) waters all points within 5 m of it. A straight path runs along the \(x\)-axis.
(a) Write down the equation of the edge of the watered region.
(b) Find the length of path that gets watered.
(c) Is a tree at \((9, 6)\) watered? Explain.
(d) The sprinkler is moved along the line \(x = 4\), away from the path, until the watered circle just touches the path. State the new equation of the edge.
Answer key and marking notes
- \((x + 3)^{2} + (y - 1)^{2} = 7\).
- \(x^{2} + 6x + 9 + y^{2} - 2y + 1 = 7 \Rightarrow x^{2} + y^{2} + 6x - 2y + 3 = 0\).
- Divide by 2: \(x^{2} + y^{2} - 4x + 6y - 3 = 0\).
- \((x - 2)^{2} - 4 + (y + 3)^{2} - 9 = 3 \Rightarrow (x - 2)^{2} + (y + 3)^{2} = 16\).
- Centre \((2, -3)\), radius 4.
- \((x - 2)^{2} + (y - 1)^{2} = 25\): centre \(C(2, 1)\).
- Gradient \(CP = \dfrac{5 - 1}{-1 - 2} = -\dfrac{4}{3}\), so the tangent gradient is \(\dfrac{3}{4}\).
- \(y - 5 = \dfrac{3}{4}(x + 1) \Rightarrow 4y - 20 = 3x + 3\).
- \(3x - 4y + 23 = 0\).
- (a) Gradient \(AC = \dfrac{4}{2} = 2\), gradient \(BC = \dfrac{4}{-8} = -\dfrac{1}{2}\); \(2 \times \left(-\dfrac{1}{2}\right) = -1\), so \(AC \perp BC\).
- (b) By the angle in a semicircle, \(AB\) is a diameter: centre \((4, 1)\), radius \(\dfrac{10}{2} = 5\).
- \((x - 4)^{2} + (y - 1)^{2} = 25\) (check \(C\): \(9 + 16 = 25\)).
- (c) \(x = 0\): \(16 + (y - 1)^{2} = 25 \Rightarrow (y - 1)^{2} = 9 \Rightarrow y = 4\) or \(y = -2\).
- The circle crosses the \(y\)-axis at \((0, 4)\) and \((0, -2)\).
- \((x - 3)^{2} + (y + 1)^{2} = 15 + 9 + 1 = 25\): centre \((3, -1)\), radius 5.
- Distance from the centre to \((7, 3)\): \(\sqrt{4^{2} + 4^{2}} = \sqrt{32}\).
- \(\sqrt{32} > \sqrt{25} = 5\), so the point is outside the circle.
- (a) \((x - 4)^{2} + (y - 3)^{2} = 25\).
- (b) \(y = 0\): \((x - 4)^{2} = 16 \Rightarrow x = 0\) or \(x = 8\); the watered length is 8 m.
- (c) Distance\(^{2}\) from \(S\) to the tree \(= 5^{2} + 3^{2} = 34 > 25\), so the tree is outside the circle and is not watered.
- (d) The path is a tangent, so the centre is 5 m above it: \(S(4, 5)\).
- New edge: \((x - 4)^{2} + (y - 5)^{2} = 25\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
State the centre and radius of the circle \((x - 4)^{2} + (y + 1)^{2} = 49\).
- ACentre \((4,\ -1)\), radius 7
- BCentre \((-4,\ 1)\), radius 7
- CCentre \((4,\ -1)\), radius 49
- DCentre \((4,\ 1)\), radius 7
Find the equation of the circle with centre \((-2,\ 3)\) that passes through the point \((1,\ 7)\).
- A\((x + 2)^{2} + (y - 3)^{2} = 25\)
- B\((x + 2)^{2} + (y - 3)^{2} = 5\)
- C\((x - 2)^{2} + (y + 3)^{2} = 25\)
- D\((x + 2)^{2} + (y - 3)^{2} = 50\)
Find the centre and radius of the circle \(x^{2} + y^{2} - 8x + 6y - 11 = 0\).
- ACentre \((4,\ -3)\), radius 6
- BCentre \((-4,\ 3)\), radius 6
- CCentre \((4,\ -3)\), radius \(\sqrt{14}\)
- DCentre \((4,\ -3)\), radius 36
The points \(A(-3,\ 4)\) and \(B(5,\ -2)\) are the ends of a diameter of a circle. Find the equation of the circle.
- A\((x - 1)^{2} + (y - 1)^{2} = 25\)
- B\((x - 1)^{2} + (y - 1)^{2} = 100\)
- C\((x - 4)^{2} + (y + 3)^{2} = 25\)
- D\((x + 1)^{2} + (y + 1)^{2} = 25\)
The point \(P(5,\ 1)\) lies on the circle \((x - 2)^{2} + (y + 3)^{2} = 25\). Find the equation of the tangent to the circle at P.
- A\(3x + 4y - 19 = 0\)
- B\(4x - 3y - 17 = 0\)
- C\(5x + y - 26 = 0\)
- D\(3x - 4y - 11 = 0\)
The equation \(x^{2} + y^{2} - 4x + 6y + k = 0\) represents a circle with radius 5. Find the value of the constant \(k\).
- A\(-12\)
- B\(12\)
- C\(38\)
- D\(8\)
A circle has its centre on the \(y\)-axis and passes through the points \((4,\ 1)\) and \((2,\ 5)\). Find the equation of the circle.
- A\(x^{2} + y^{2} - 3y - 14 = 0\)
- B\(x^{2} + y^{2} - 3x - 14 = 0\)
- C\(x^{2} + y^{2} + 3y - 20 = 0\)
- D\(x^{2} + y^{2} - 3y - 16.25 = 0\)
Find the shortest distance from the point \(P(8,\ 3)\) to the circle \(x^{2} + y^{2} - 6x + 2y - 15 = 0\).
- A\(\sqrt{41} - 5\)
- B\(\sqrt{41}\)
- C\(\sqrt{41} + 5\)
- D\(5\sqrt{5} - 5\)
Quiz key
\((x - 4)^{2}\) gives \(a = 4\), and \((y + 1)^{2} = \big(y - (-1)\big)^{2}\) gives \(b = -1\). \(r^{2} = 49\), so \(r = 7\). The tempting \((-4,\ 1)\) reads the signs in the brackets literally. Radius 49 forgets that the right-hand side is \(r^{2}\), and \((4,\ 1)\) gets only one sign wrong.
\(r^{2} = (1 + 2)^{2} + (7 - 3)^{2} = 9 + 16 = 25\). With centre \((-2,\ 3)\): \((x + 2)^{2} + (y - 3)^{2} = 25\). The tempting \(= 5\) puts \(r\) instead of \(r^{2}\) on the right. \((x - 2)^{2} + (y + 3)^{2}\) has centre \((2,\ -3)\), and 50 is the squared distance of \((1,\ 7)\) from the ORIGIN, not from the centre.
\((x - 4)^{2} - 16 + (y + 3)^{2} - 9 - 11 = 0\), so \((x - 4)^{2} + (y + 3)^{2} = 36\). Centre \((4,\ -3)\), radius 6. The tempting \((-4,\ 3)\) reads the signs of \(-8x\) and \(+6y\) directly instead of halving and changing sign. \(\sqrt{14}\) comes from \(16 + 9 - 11\), with the wrong sign on the constant, and 36 is \(r^{2}\).
Midpoint \(= \left(\dfrac{-3 + 5}{2},\ \dfrac{4 - 2}{2}\right) = (1,\ 1)\). \(\text{AB} = \sqrt{8^{2} + 6^{2}} = 10\), so \(r = 5\) and \(r^{2} = 25\): \((x - 1)^{2} + (y - 1)^{2} = 25\). The tempting 100 squares the diameter instead of the radius. \((4,\ -3)\) is half the CHANGE from A to B, not the midpoint, and \((x + 1)^{2} + (y + 1)^{2}\) has centre \((-1,\ -1)\).
C is \((2,\ -3)\). Gradient of CP \(= \dfrac{1 + 3}{5 - 2} = \dfrac{4}{3}\), so the tangent has gradient \(-\dfrac{3}{4}\): \(y - 1 = -\dfrac{3}{4}(x - 5)\), giving \(4y - 4 = -3x + 15\), i.e. \(3x + 4y - 19 = 0\). The tempting \(4x - 3y - 17 = 0\) is the line CP itself (the normal). \(5x + y - 26 = 0\) is perpendicular to OP, using the origin instead of the centre, and \(3x - 4y - 11 = 0\) uses \(+\frac{3}{4}\), forgetting the minus sign.
\((x - 2)^{2} - 4 + (y + 3)^{2} - 9 + k = 0\), so \((x - 2)^{2} + (y + 3)^{2} = 13 - k\). Then \(13 - k = 25\), giving \(k = -12\). The tempting 12 is a sign slip in the last step. 38 solves \(k - 13 = 25\), moving the constants with the wrong signs, and 8 sets \(13 - k = 5\), using \(r\) instead of \(r^{2}\).
\(4^{2} + (1 - b)^{2} = 2^{2} + (5 - b)^{2}\) gives \(17 - 2b = 29 - 10b\), so \(b = \dfrac{3}{2}\). \(r^{2} = 16 + \left(\dfrac{1}{2}\right)^{2} = 16.25\). The equation \(x^{2} + \left(y - \dfrac{3}{2}\right)^{2} = 16.25\) expands to \(x^{2} + y^{2} - 3y + 2.25 - 16.25 = 0\), i.e. \(x^{2} + y^{2} - 3y - 14 = 0\). Check \((2,\ 5)\): \(4 + 25 - 15 - 14 = 0\). The tempting \(-16.25\) forgets the \(+2.25\) from expanding the bracket. \(-3x\) puts the centre on the \(x\)-axis, and \(+3y - 20\) uses the centre \(\left(0,\ -\frac{3}{2}\right)\), which fits \((4,\ 1)\) but not \((2,\ 5)\).
\((x - 3)^{2} + (y + 1)^{2} = 9 + 1 + 15 = 25\): centre \(C(3,\ -1)\), radius 5. \(\text{CP} = \sqrt{5^{2} + 4^{2}} = \sqrt{41} > 5\), so P is outside the circle, and the shortest distance is \(\sqrt{41} - 5 \approx 1.40\). The tempting \(\sqrt{41}\) is the distance to the CENTRE. \(\sqrt{41} + 5\) is the greatest distance to the circle, and \(5\sqrt{5} - 5\) uses the centre \((-3,\ 1)\), with the signs read wrongly.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.