9709 Pure 1 · 1.3 Coordinate Geometry · Session 2 of 4

The Equation of a Circle

Today a circle becomes an equation you can read: where its center is, how big it is, and which way the tangent goes. You met circle equations on Khan Academy. Today is about doing them the way Cambridge marks them.

Do Now, and why circles matter7 min
The circle equation, how to read it, and the Circle Reader10 min
The expanded form, a route check, and inside, on or outside14 min
The tangent at a point, the Tangent Lab, and a 3-mark start from June 202314 min
Spot the mistake, practice together, then on your own8 min
Recap and exit ticket5 min

Do Now · 5 min · no notes, no calculator

\(C(1, -2)\) and \(P(4, 4)\). Find the distance \(CP\), the gradient of \(CP\), and the gradient of a line perpendicular to \(CP\).

Give the distance in exact form. Sketch the two points first.

Write all three, then tap to check.
distance:  \(\sqrt{(4 - 1)^2 + (4 - (-2))^2} = \sqrt{9 + 36} = \sqrt{45} = \sqrt{9}\sqrt{5} = 3\sqrt{5}\)
gradient:  \(\dfrac{4 - (-2)}{4 - 1} = \dfrac{6}{3} = 2\)
perpendicular gradient, flip and change the sign:  \(-\dfrac{1}{2}\),  check \(2 \times \left(-\tfrac{1}{2}\right) = -1\)
\(CP = 3\sqrt{5}\), gradient \(2\), perpendicular gradient \(-\tfrac{1}{2}\). All three come back today.

Why this matters · 2 min

A circle is every point that is the same distance from one fixed point, the center.

That distance is the radius. Turn "the same distance" into algebra and you get one equation that holds for every point on the circle and for no other point.

What each skill answers

Center and radius: where is the middle, and how far does it reach?

Inside, on or outside: is this place covered or not?

Tangent: which way does something go the moment it leaves the circle?

Where you will see it

Hurricane wind fields. Forecasters give how far out from the eye the storm-force winds reach. Is St. Croix inside that circle or outside it?

Phones and Wi-Fi. A tower or router covers every point within a certain distance. That coverage area is a circle.

GPS. Your phone knows how far it is from each satellite. Each distance is a circle (a sphere, in space), and you are where they meet.

Sprinklers and radar. Both sweep out a circle around a fixed center.

Letting go. A ball whirled on a string and released flies off along the tangent, at right angles to the string.

Your Do Now distance was a radius, and your perpendicular gradient was a tangent. By the end you can do both from an equation, which is how a real Cambridge question starts.

Direct instruction · 1 of 13 · where it comes from

Every point \(3\sqrt{5}\) from \(C(1, -2)\). What equation do they all satisfy?

radius \(CP\)   across and up   the circle

Take any point \((x, y)\) on the circle. Its distance from \(C(1, -2)\) is the radius.

across is \(x - 1\), up is \(y - (-2) = y + 2\):
\(\sqrt{(x - 1)^2 + (y + 2)^2} = 3\sqrt{5}\)
square both sides, and \((3\sqrt{5})^2 = 9 \times 5 = 45\):
\((x - 1)^2 + (y + 2)^2 = 45\)

Check with \(P(4, 4)\): \((4 - 1)^2 + (4 + 2)^2 = 9 + 36 = 45\). It works. That is your Do Now, squared.

Center \((a, b)\), radius \(r\):
\((x - a)^2 + (y - b)^2 = r^2\)

This is Pythagoras with the square root squared away. It is not in the formula booklet, so you need it from memory.

Direct instruction · 2 of 13 · reading the equation

Read the center with the signs flipped. Read the radius as the square root of the right-hand side.

\((x + 3)^2 + (y - 5)^2 = 49\)
\(x + 3 = x - (-3)\):  \(a = -3\),  \(b = 5\)
center \((-3, 5)\), radius \(\sqrt{49} = 7\)
\((x - 2)^2 + y^2 = 20\)
\(y^2 = (y - 0)^2\):  \(a = 2\),  \(b = 0\)
center \((2, 0)\), radius \(\sqrt{20} = 2\sqrt{5}\)
\(x^2 + y^2 = 16\)
both brackets are \((x - 0)\) and \((y - 0)\):
center \((0, 0)\), radius \(4\)

Three mistakes from real exam scripts (June 2023 examiner report)

Radius on the right instead of radius squared. A circle with radius 10 is \(\ldots = 100\). Writing "= 10" or "= \(\sqrt{10}\)" was common.

The sign of the center. \((x + 3)^2\) means \(a = -3\). The center is \((-3, 5)\), not \((3, 5)\).

\(x\) and \(y\) swapped. The first number in the center goes with \(x\). Also watch for brackets that were never squared.

Direct instruction · 3 of 13 · the Circle Reader

Set the equation, pick a reading, predict, then show it.

the real circle for this equation    a reading that matches    a reading that misses

Direct instruction · 4 of 13 · the expanded form

Find the center and radius of \(x^2 + y^2 + 10x - 4y - 7 = 0\).

Multiply out the brackets and a circle can look like this. Two routes get you back.

Route 1 · complete the square

group the \(x\) terms and the \(y\) terms:
\((x^2 + 10x) + (y^2 - 4y) - 7 = 0\)
half of 10 is 5, half of \(-4\) is \(-2\):
\((x + 5)^2 - 25 + (y - 2)^2 - 4 - 7 = 0\)

Why subtract 25? \((x + 5)^2\) expands to \(x^2 + 10x + 25\). That is 25 more than you started with, so take it back off. Same for \((y - 2)^2\) and the 4.

move the numbers across, \(25 + 4 + 7 = 36\):
\((x + 5)^2 + (y - 2)^2 = 36\)
center \((-5, 2)\), radius \(6\)

Route 2 · match the pattern

\(x^2 + y^2 + 2gx + 2fy + c = 0\) has center \((-g, -f)\) and radius \(\sqrt{g^2 + f^2 - c}\)
match:  \(2g = 10\), so \(g = 5\);  \(2f = -4\), so \(f = -2\);
\(c = -7\)
center \((-g, -f)\):  \((-5, 2)\)
radius:  \(\sqrt{25 + 4 - (-7)} = \sqrt{36} = 6\)
Same answer: center \((-5, 2)\), radius \(6\)

Route 2 is Route 1 done once with letters. The next slide shows where it comes from.

Direct instruction · 5 of 13 · where the pattern comes from

Route 2 is Route 1 with letters in place of the numbers.

Write the \(x\) coefficient as \(2g\) and the \(y\) coefficient as \(2f\). Half of \(2g\) is just \(g\), which is why the 2 is there. Left: any circle. Right: the one we just did.

Any circle

\(x^2 + y^2 + 10x - 4y - 7 = 0\), so \(g = 5\), \(f = -2\), \(c = -7\)

1 · group the \(x\) terms and the \(y\) terms:
\((x^2 + 2gx) + (y^2 + 2fy) + c = 0\)
\((x^2 + 10x) + (y^2 - 4y) - 7 = 0\)
2 · complete each square. Half of \(2g\) is \(g\), and you take \(g^2\) back off:
\((x + g)^2 - g^2 + (y + f)^2 - f^2 + c = 0\)
\((x + 5)^2 - 25 + (y - 2)^2 - 4 - 7 = 0\)
3 · move the numbers to the right-hand side:
\((x + g)^2 + (y + f)^2 = g^2 + f^2 - c\)
\((x + 5)^2 + (y - 2)^2 = 25 + 4 + 7 = 36\)
4 · read it like slide 5: the center makes each bracket zero, and the right-hand side is \(r^2\):
center \((-g, -f)\),  radius \(\sqrt{g^2 + f^2 - c}\)
center \((-5, 2)\),  radius \(\sqrt{36} = 6\)

So the pattern is not a new rule. It is the answer to step 3, written once for every circle. If \(g^2 + f^2 - c\) is negative, there is no square root, and that is the "not a circle" case on the next slide.

Direct instruction · 6 of 13 · two checks before you start

Is the equation ready, and is it really a circle?

Check 1 · \(x^2\) and \(y^2\) must have coefficient 1

\(2x^2 + 2y^2 - 8x + 12y + 8 = 0\)
divide every term by 2 first:
\(x^2 + y^2 - 4x + 6y + 4 = 0\)
complete the square:
\((x - 2)^2 - 4 + (y + 3)^2 - 9 + 4 = 0\)
\(4 + 9 - 4 = 9\):  \((x - 2)^2 + (y + 3)^2 = 9\)
center \((2, -3)\), radius \(3\)

This is the same first move as dividing through by \(a\) when you completed the square in 1.1.

Check 2 · the right-hand side must be positive

\(x^2 + y^2 + 2x + 4y + 10 = 0\)
\((x + 1)^2 - 1 + (y + 2)^2 - 4 + 10 = 0\)
\(1 + 4 - 10 = -5\):  \((x + 1)^2 + (y + 2)^2 = -5\)

Two squares added together can never be negative, so no point fits. This is not a circle.

Also check the squared terms. Expand \((x - a)^2 + (y - b)^2\) and you always get \(x^2\) and \(y^2\) with the same coefficient, and \(x\) is never multiplied by \(y\). So \(x^2 + 2y^2 + \ldots\) is not a circle (it is an oval), and neither is anything with an \(xy\) term (the shape is tilted).

Direct instruction · 7 of 13 · route check

Route check. Do not solve these.

Look at each equation and say what it needs first. That is the whole task. Ten seconds each.

1.  \((x - 4)^2 + (y + 1)^2 = 12\)
Nothing. Read it now: center \((4, -1)\), radius \(\sqrt{12} = 2\sqrt{3}\)
2.  \(x^2 + y^2 - 8x + 2y = 0\)
Complete the square. \((x - 4)^2 + (y + 1)^2 = 17\)
3.  \(3x^2 + 3y^2 + 6x - 12y - 12 = 0\)
Divide by 3 first, then complete the square. \((x + 1)^2 + (y - 2)^2 = 9\)
4.  \(x^2 + 2y^2 + 4x - 6 = 0\)
Stop. \(x^2\) and \(y^2\) have different coefficients, so this is not a circle.

Which route for the expanded form?

Completing the square always works and shows every step, so it is the safe one to write in an exam. The pattern \((-g, -f)\), \(\sqrt{g^2 + f^2 - c}\) is a fast check. It only works once \(x^2\) and \(y^2\) have coefficient 1, and it is not in the formula booklet.

Numbers 1 and 2 have the same center. The only difference is whether the squares are already done for you.

Direct instruction · 8 of 13 · inside, on or outside

Put the point into the left-hand side and compare with \(r^2\).

The circle from the start: \((x - 1)^2 + (y + 2)^2 = 45\)
\((5, 1)\)
\((5 - 1)^2 + (1 + 2)^2 = 16 + 9 = 25\)
\(25 < 45\): inside
\((7, 1)\)
\((7 - 1)^2 + (1 + 2)^2 = 36 + 9 = 45\)
\(45 = 45\): on the circle
\((-5, 2)\)
\((-5 - 1)^2 + (2 + 2)^2 = 36 + 16 = 52\)
\(52 > 45\): outside

Why this works: the left-hand side is the distance from the center, squared. Less than \(r^2\) means closer than the radius. No square roots needed.

Direct instruction · 9 of 13 · hurricanes

Is the town inside the storm-force winds?

edge of the storm-force winds   one grid square = 10 km

A made-up storm. Put Christiansted at the origin, one grid unit = 10 km. The eye is at \((6, 8)\), and storm-force winds reach 120 km from the eye, so \(r = 12\).

the edge of the winds:  \((x - 6)^2 + (y - 8)^2 = 144\)
Christiansted \((0, 0)\):  \(36 + 64 = 100\), and \(100 < 144\)
Inside. Christiansted is 100 km from the eye.
Frederiksted, about \((-2, -1)\):  \(64 + 81 = 145\), and \(145 > 144\)
Outside, but only just. \(\sqrt{145} \approx 12.04\), so about 120.4 km away.

Same test as the last slide, with real places. Half a kilometer decides it, which is why forecasters care about the exact radius.

Hurricane Closest Approach: how close does a moving storm come?

Direct instruction · 10 of 13 · tangent at a point

A tangent is perpendicular to the radius at the point where it touches.

radius \(CP\)   circle   tangent at \(P\)

Find the tangent to \((x - 1)^2 + (y + 2)^2 = 45\) at \(P(4, 4)\). You need a point (you have \(P\)) and a gradient.

1 · center from the equation:  \(C(1, -2)\)
2 · gradient of the radius \(CP\):  \(\dfrac{4 - (-2)}{4 - 1} = 2\)
3 · flip and change the sign:  tangent gradient \(-\tfrac{1}{2}\)
4 · line through \(P\):  \(y - 4 = -\tfrac{1}{2}(x - 4)\)
times 2:  \(2y - 8 = -x + 4\)
\(x + 2y - 12 = 0\)

Why a right angle? Tilt the line even a little away from \(90^\circ\) and one side of it dips inside the circle, so it crosses the circle again. Only the line at right angles to the radius stays outside everywhere except \(P\). The Tangent Lab shows this next.

The trap: using the radius gradient, 2, as the tangent gradient. The radius goes into the circle. The tangent only touches it. You do not need calculus for this, and the examiners say students who tried it did worse.

Direct instruction · 11 of 13 · the Tangent Lab

Pick a point and a method. Does the line only touch, or does it cut through?

radius    a tangent: touches at \(P\) only    a line that cuts the circle again

Direct instruction · 12 of 13 · June 2023, Paper 11, Q12

A real Cambridge question opens exactly like today's lesson.

The diagram shows a circle \(P\) with center \((0, 2)\) and radius 10 and the tangent to the circle at the point \(A\) with coordinates \((6, 10)\). It also shows a second circle \(Q\) with center at the point where this tangent meets the \(y\)‑axis and with radius \(\tfrac{5}{2}\sqrt{5}\).

(a) Write down the equation of circle \(P\). [1]

(b) Find the equation of the tangent to the circle \(P\) at \(A\). [2]

Parts (c) and (d), where the tangent meets circle \(Q\), are next lesson.

Sketch first: the center is on the \(y\)-axis and \(A\) is up and to the right of it, so the radius slopes up and the tangent must slope down.

circle \(P\)   circle \(Q\)   tangent at \(A\)

Direct instruction · 13 of 13 · Q12(a)(b) · where the 3 marks come from

Three marks, three lines of working.

(a) B1 · the equation, read straight from the center and radius:
\(x^2 + (y - 2)^2 = 100\)
earn it: \(-2\) inside the bracket, and \(10^2 = 100\) on the right
(b) M1 · the gradient of the radius, or of the tangent:
\(m_{\text{radius}} = \dfrac{10 - 2}{6 - 0} = \dfrac{4}{3}\),  so \(m_{\text{tangent}} = -\dfrac{3}{4}\)
earn it: either gradient seen
(b) A1 · the tangent through \(A\):
\(y - 10 = -\tfrac{3}{4}(x - 6)\),  so \(y = -\tfrac{3}{4}x + \tfrac{29}{2}\)
earn it: any correct form. This question does not ask for \(ax + by + c = 0\).
\(y = -\tfrac{3}{4}x + \tfrac{29}{2}\),  or \(3x + 4y - 58 = 0\)

Reading the codes

B1 is a mark for a correct statement that needs no method. Here, writing the equation down.

M1 then A1: a method mark, then an accuracy mark that needs it. There are no stars or D marks in this part.

What the examiners saw

In (a), "= 10" and "= \(\sqrt{10}\)" were common, along with unsquared brackets, swapped \(x\) and \(y\), and sign errors.

In (b), some students turned the gradient upside down without changing the sign, or forgot to take the perpendicular at all. Those who differentiated did worse.

Check: \(A(6, 10)\) must be on the answer. \(3(6) + 4(10) - 58 = 0\). It works.

Spot the mistake · 3 min

Three answers to that question. What is wrong with each? Tap to check.

Circle \(P\): center \((0, 2)\), radius 10. Tangent at \(A(6, 10)\).
A.  "(a) \(x^2 + (y - 2)^2 = 10\)"
test \(A(6, 10)\), which is on the circle:  \(36 + 64 = 100\), not 10
The right-hand side is \(r^2\). It should be \(= 100\).
B.  "(a) \(x^2 + (y + 2)^2 = 100\)"
read it back:  \((y + 2)\) means \(b = -2\), so the center would be \((0, -2)\)
Wrong sign. Center \((0, 2)\) gives \((y - 2)^2\).
C.  "(b) gradient \(\tfrac{4}{3}\), so \(y - 10 = \tfrac{4}{3}(x - 6)\),  \(y = \tfrac{4}{3}x + 2\)"
put \(x = 0\):  \(y = 2\). This line goes through the center \((0, 2)\).
That is the radius, not the tangent. Flip and change the sign: \(-\tfrac{3}{4}\).

These are the mistakes the June 2023 examiner report describes for this question.

Practice · together · 5 min

Sketch first. Exact answers. Tap to check.

1.  Specimen Paper 1, Q10(a). The circle \(x^2 + y^2 + 4x - 2y - 20 = 0\) has center \(C\). State the coordinates of \(C\). [1]  Then find the radius.
\((x + 2)^2 - 4 + (y - 1)^2 - 1 - 20 = 0\)
\(4 + 1 + 20 = 25\):  \((x + 2)^2 + (y - 1)^2 = 25\)
\(C(-2, 1)\) (B1),  radius \(5\)
2.  \(x^2 + y^2 - 10x + 4y + 4 = 0\). Find the center and radius. Is \((8, 1)\) inside, on or outside?
\((x - 5)^2 - 25 + (y + 2)^2 - 4 + 4 = 0\),  so \((x - 5)^2 + (y + 2)^2 = 25\)
\((8, 1)\):  \((8 - 5)^2 + (1 + 2)^2 = 9 + 9 = 18\), and \(18 < 25\)
center \((5, -2)\), radius \(5\);  inside
3.  A circle has center \((2, -1)\) and passes through \((5, 1)\). Find its equation, and the tangent at \((5, 1)\) as \(ax + by + c = 0\).
\(r^2\):  \((5 - 2)^2 + (1 + 1)^2 = 9 + 4 = 13\),  so \((x - 2)^2 + (y + 1)^2 = 13\)
radius gradient  \(\tfrac{1 - (-1)}{5 - 2} = \tfrac{2}{3}\),  tangent  \(-\tfrac{3}{2}\)
\(y - 1 = -\tfrac{3}{2}(x - 5) \Rightarrow 2y - 2 = -3x + 15\)
\(3x + 2y - 17 = 0\)

Practice · on your own · 4 min, finish at home

Your turn. Tap to check when you have a final answer.

4.  Find the center and radius of \(2x^2 + 2y^2 + 12x - 8y - 6 = 0\).
divide by 2:  \(x^2 + y^2 + 6x - 4y - 3 = 0\)
\((x + 3)^2 - 9 + (y - 2)^2 - 4 - 3 = 0\),  so \((x + 3)^2 + (y - 2)^2 = 16\)
center \((-3, 2)\), radius \(4\)
5.  \(A(-6, 2)\) and \(B(2, 4)\) are the ends of a diameter. Find the equation of the circle, and the tangent at \(B\).
center = midpoint:  \(\left(\tfrac{-6 + 2}{2}, \tfrac{2 + 4}{2}\right) = (-2, 3)\)
\(r^2\), center to \(B\):  \(4^2 + 1^2 = 17\),  so \((x + 2)^2 + (y - 3)^2 = 17\)
radius gradient  \(\tfrac{4 - 3}{2 - (-2)} = \tfrac{1}{4}\),  tangent  \(-4\):  \(y - 4 = -4(x - 2)\)
\((x + 2)^2 + (y - 3)^2 = 17\);  \(4x + y - 12 = 0\)
6.  Find the center and radius of \(x^2 + y^2 - 8y - 9 = 0\), then the tangent at \((5, 4)\).
\(x^2 + (y - 4)^2 - 16 - 9 = 0\),  so \(x^2 + (y - 4)^2 = 25\): center \((0, 4)\), radius \(5\)
radius from \((0, 4)\) to \((5, 4)\) is horizontal,  gradient 0. You cannot flip 0, so draw it: the tangent is vertical.
\(x = 5\)

Finished early? Show that \(x^2 + y^2 - 6x + 2y + 12 = 0\) is not a circle. Then try this: a radius has gradient \(m\). Explain why the tangent has gradient \(-\tfrac{1}{m}\).

Recap

Four things to carry out of here.

\((x - a)^2 + (y - b)^2 = r^2\) is the distance formula, squared. Center \((a, b)\), signs flipped. The right-hand side is \(r^2\).

Expanded form: complete the square on \(x\) and on \(y\). Divide first if \(x^2\) and \(y^2\) do not have coefficient 1. A negative right-hand side means no circle.

Inside, on or outside: substitute the point and compare with \(r^2\).

Tangent at a point: radius gradient, then flip and change the sign. A tangent is perpendicular to the radius, so it touches the circle once.

Exit ticket · 5 min

Schoology, Paper 1 → 1.3 Coordinate Geometry.

Four questions. Do the working on paper first. For the last one, give the equation in the form \(ax + by + c = 0\) with integers.

Submit before you leave. An attempt left open scores nothing.

Homework · due Monday, October 26

Reading and workbook, as posted in Schoology.

Stuck on a question? Write down where you got stuck and bring it. That still counts as done.

Intersection Lab: slide a line across a circle. A preview of next lesson  Line Lab: check any tangent equation step by step