9709 Pure 1 · 1.3 Coordinate Geometry · Booklet 2 of 3
Circles
A circle is one idea, every point the same distance from a center, turned into one equation. This booklet shows you how to read that equation, how to rebuild it when it has been multiplied out, how to test any point against it, and how to find the tangent at a point. It ends with a real Cambridge question, marked line by line.
By the end you can
- write a circle's equation from its center and radius
- read the center and radius from any circle equation
- say whether a point is inside, on or outside a circle
- find the tangent to a circle at a point
You need
- distance, gradient and perpendicular gradient (Booklet 1, Straight lines)
- completing the square (1.1 Quadratics)
How to use it
Read in order. When a box asks you to predict, decide before you click. Do each Check yourself question on paper first, then mark yourself. Your progress is saved in this browser.
A circle is a distance
Where the equation comes from, so you never have to memorize it blind.
The idea
A circle is every point that is the same distance from one fixed point. The fixed point is the center. The distance is the radius.
\(C(1, -2)\) and \(P(4, 4)\). Find the distance \(CP\) in exact form, the gradient of \(CP\), and the gradient of a line perpendicular to \(CP\).
Keep these three answers. The distance is about to become a radius, and the perpendicular gradient becomes a tangent in section 6.
Every point \((x, y)\) on this circle is \(\sqrt{45}\) from \(C(1, -2)\). Write that down as an equation.
That is the whole idea: the circle's equation is the distance formula with the square root squared away.
Try it · the Circle Probe
Drag the point. The left-hand side is worked out for wherever you put it.
Find three more points where the total is exactly 45. (There are eight whole-number points on this circle. Here is a hint: \(45 = 36 + 9\).)
The equation of a circle
This is not in the formula booklet (MF19). Learn it from the picture above, so you can always rebuild it.
Reading center and radius
Two rules, and the three mistakes Cambridge examiners see most.
The two rules
Center: the values of \(x\) and \(y\) that make each bracket zero. In \((x + 3)^2\), \(x + 3 = 0\) gives \(x = -3\). So the signs look flipped.
Radius: the square root of the right-hand side. The right-hand side is \(r^2\), not \(r\).
center
What is the center of \((x + 3)^2 + (y - 5)^2 = 49\)?
radius
What is the radius of \((x - 2)^2 + y^2 = 20\)?
Going the other way
A circle has center \((2, -5)\) and radius 3. Put the center into the brackets with the signs flipped, and square the radius:
Try it · the Circle Reader
Set an equation, pick a way of reading it, predict, then show the circle that reading draws.
the real circle a reading that matches a reading that misses
What examiners saw (June 2023)
Radius, not radius squared
Radius 10, center \((0, 2)\):
The right-hand side is \(r^2 = 100\).
Signs kept as written
From \((x + 3)^2 + (y - 5)^2 = 49\):
Make each bracket zero: \((-3, 5)\).
\(x\) and \(y\) swapped
Center \((4, -1)\):
The first number goes with \(x\): \((x - 4)^2 + (y + 1)^2\).
The examiners also saw brackets that were never squared. A circle always has both brackets squared.
The expanded form
When the brackets have been multiplied out, complete the square to get them back.
Multiply out \((x + 5)^2 + (y - 2)^2 = 36\) and you get \(x^2 + 10x + 25 + y^2 - 4y + 4 = 36\), which tidies up to
Same circle, but you can no longer read the center. Cambridge often gives circles this way. Here are two routes back.
Route 1 · complete the square (always works)
Why subtract 25?
\((x + 5)^2 = x^2 + 10x + 25\). That is 25 more than \(x^2 + 10x\), so you take the 25 back off. Same for \((y - 2)^2\) and the 4. This is the same move as completing the square in 1.1.
Route 2 · Route 1 with letters
Write any expanded circle as \(x^2 + y^2 + 2gx + 2fy + c = 0\). The 2 is there because half of \(2g\) is just \(g\). Now do Route 1 once, with letters on the left and our numbers on the right (\(g = 5\), \(f = -2\), \(c = -7\)).
Which route?
Write Route 1 in an exam. It always works and shows every step for the method mark. Use Route 2 as a fast check. It only works once \(x^2\) and \(y^2\) have coefficient 1, and it is not in the formula booklet.
center
What is the center of \(x^2 + y^2 - 6x + 8y = 0\)?
Is it really a circle?
Look before you start: is the equation ready, and could it be a circle at all?
Check 1 · \(x^2\) and \(y^2\) need coefficient 1
Check 2 · the right-hand side must be positive
Two squares added together are never negative, so no point fits. This is not a circle. In Route 2 terms, \(g^2 + f^2 - c\) is negative, so there is no square root.
Why a circle has equal \(x^2\) and \(y^2\), and no \(xy\)
Expand any \((x - a)^2 + (y - b)^2\). Each bracket gives exactly one \(x^2\) or one \(y^2\), so the two always match. And \(x\) only lives in the first bracket and \(y\) in the second, so they are never multiplied together.
So \(x^2 + 2y^2 + \ldots\) is not a circle. The unequal coefficients stretch it into an oval. Anything with an \(xy\) term is tilted, so it is not a circle either.
Route check. Do not solve these.
Say what each one needs first. That is the whole task.
first move
\((x - 4)^2 + (y + 1)^2 = 12\)
first move
\(x^2 + y^2 - 8x + 2y = 0\)
first move
\(3x^2 + 3y^2 + 6x - 12y - 12 = 0\)
first move
\(x^2 + 2y^2 + 4x - 6 = 0\)
Inside, on or outside
Put the point in, compare with \(r^2\). No square roots needed.
The test
Substitute the point into the left-hand side of \((x - a)^2 + (y - b)^2 = r^2\). Less than \(r^2\): inside. Equal: on. More: outside.
Why it works
The left-hand side is the distance from the center, squared. Comparing squares is the same as comparing the distances, because distances are never negative. That is why you never need the square root.
On our circle, \((x - 1)^2 + (y + 2)^2 = 45\)
Watch the signs: \((-5 - 1)^2 = (-6)^2 = 36\). The square makes it positive. Go back to the Circle Probe in section 1 and check all three.
St. Croix · is the town inside the storm?
A made-up storm. Put Christiansted at the origin, with one grid unit = 10 km. The eye is at \((6, 8)\), and storm-force winds reach 120 km from the eye, so \(r = 12\) and the edge of the winds is \((x - 6)^2 + (y - 8)^2 = 144\).
Predict first: is Christiansted inside? Is Frederiksted, at about \((-2, -1)\)? Then use the buttons to check.
Christiansted gives \(36 + 64 = 100 < 144\): inside, 100 km from the eye. Frederiksted gives \(64 + 81 = 145 > 144\): outside, but only just (about 120.4 km). Half a kilometer decides it, which is why forecasters care about the exact radius.
The tangent at a point
A tangent is perpendicular to the radius where it touches.
The fact
The tangent at \(P\) is at right angles to the radius \(CP\). So: find the gradient of the radius, flip it and change the sign, and draw the line through \(P\).
Why a right angle?
Tilt a line through \(P\) even a little away from \(90^\circ\) and one side of it dips inside the circle, so it crosses the circle a second time. Only the line at right angles to the radius stays outside everywhere except at \(P\). The Tangent Lab below lets you see this.
Tangent to \((x - 1)^2 + (y + 2)^2 = 45\) at \(P(4, 4)\)
These are your warm-up answers from section 1: the radius gradient was 2 and the perpendicular gradient was \(-\tfrac{1}{2}\).
Try it · the Tangent Lab
Pick a point and a method. Does the line only touch, or does it cut through?
The trap
Tangent gradient \(= 2\), the radius gradient.
The radius goes into the circle through the center. The tangent only touches it. Flip and change the sign. You do not need calculus here, and the examiners say students who tried it did worse.
a special case
The circle \(x^2 + y^2 = 25\) passes through \((0, 5)\). What is the tangent there?
Exam corner
A real Cambridge question opens exactly like this booklet. Try it, then watch the marks light up.
June 2023 · Paper 11 · Question 12 (a) and (b)
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 in Booklet 3.
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. Now write (a) and (b) on paper before you go on.
Check: \(A(6, 10)\) must be on the answer. \(3(6) + 4(10) - 58 = 0\). It works.
Reading the codes
M is a method mark: a valid method with the numbers put in. A is an accuracy mark for the correct answer, and it needs the M before it. B is a mark for a correct fact or value that needs no method. A star and a D (as in Booklet 1) mean a later mark depends on an earlier one. This part has none.
Find the mistake
Three answers to this question. Decide what is wrong with each, then open it.
Answer A
Test \(A(6, 10)\), which is on the circle: \(36 + 64 = 100\), not 10. The right-hand side is \(r^2\).
Answer B
\((y + 2)\) is zero at \(y = -2\), so this circle's center is \((0, -2)\). Center \((0, 2)\) needs \((y - 2)^2\).
Answer C
Put \(x = 0\): \(y = 2\). This line goes through the center, so it is the radius. Flip and change the sign: \(-\tfrac{3}{4}\).
These are the mistakes the June 2023 examiner report describes for this question. Answer C's check, "does my tangent go through the center?", catches the error every time.
Check yourself
Work on paper, open the answer, then mark yourself. Exact answers, sketch first.
The circle \(x^2 + y^2 + 4x - 2y - 20 = 0\) has center \(C\). State the coordinates of \(C\). [1] Then find the radius.
Find the center and radius of \(x^2 + y^2 - 10x + 4y + 4 = 0\). Is \((8, 1)\) inside, on or outside it?
A circle has center \((-3, 4)\) and passes through the origin. Find its equation.
Find the center and radius of \(2x^2 + 2y^2 + 12x - 8y - 6 = 0\).
The circle \(x^2 + y^2 - 6x + 2y + k = 0\) has radius 4. Find \(k\).
The point \((a, 5)\) lies on the circle \((x - 2)^2 + (y - 1)^2 = 25\). Find the two possible values of \(a\).
Both answers count. A horizontal line meets a circle twice.
A circle has center \((2, -1)\) and passes through \((5, 1)\). Find its equation, and the tangent at \((5, 1)\) in the form \(ax + by + c = 0\).
\(A(-6, 2)\) and \(B(2, 4)\) are the ends of a diameter. Find the equation of the circle, and the tangent at \(B\).
Find the center and radius of \(x^2 + y^2 - 8y - 9 = 0\), then the tangent at \((5, 4)\).
On a plan of the school, one unit is 5 m. The Wi-Fi router is at \((2, 3)\) and reaches 30 m in every direction. Is a classroom door at \((7, 6)\) covered?
Show that \(x^2 + y^2 - 6x + 2y + 12 = 0\) is not a circle.
A radius has gradient \(m\), where \(m \ne 0\). Explain why the tangent at the end of that radius has gradient \(-\dfrac{1}{m}\).
The tangent is perpendicular to the radius. Perpendicular gradients multiply to \(-1\), so \(m \times m_{\text{tangent}} = -1\), which gives \(m_{\text{tangent}} = -\dfrac{1}{m}\). Say it in words in an exam: "the product of the gradients is \(-1\)."
Four things to carry with you
- \((x - a)^2 + (y - b)^2 = r^2\) is the distance formula, squared. The center makes each bracket zero. The right-hand side is \(r^2\).
- Expanded form: divide first if \(x^2\) and \(y^2\) do not have coefficient 1, then complete the square on \(x\) and on \(y\). 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. Sketch first when a gradient is 0 or undefined.
Next · Booklet 3
Where lines meet curves
Where a line meets a circle, how to tell if it cuts, touches or misses using the discriminant from 1.1, and parts (c) and (d) of the June 2023 question.