Co-ordinate Geometry · Paper 2
The Circle 1
Higher Level · Circle centred at the origin · Tap NEXT to begin
Section 1 of 9
The Parts of a Circle
A circle is defined by two things only: its centre and its radius. Everything else we say about circles is built on those two.
Before any equations, learn the names of the parts. You will be expected to use them naturally in answers.
Words you must know
1Radius — a line from the centre to the curve. (Plural: radii.)
2Chord — a straight line joining any two points on the curve.
3Diameter — a chord that passes through the centre. Length $= 2r$.
4Tangent — a line that touches the circle at exactly one point. It is perpendicular to the radius at that point.
5Sector — a "slice of pie": two radii and the arc between them.
6Segment — the region cut off by a chord (the area between a chord and the arc).
7Circumference — the curve itself, the perimeter of the circle. Length $= 2\pi r$.
Key fact to keep in mind for everything that follows: the tangent is perpendicular to the radius drawn to the point of contact. This single fact powers nearly every tangent problem in this lesson.
You try
A line cuts a circle at two points but does not pass through the centre. What is that line called?
It joins two points on the curve — but it isn't a diameter (no centre).
Two points on the curve $\Rightarrow$ chord or diameter.
Doesn't pass through the centre $\Rightarrow$ not a diameter.
It's a chord.
Chord
Section 2 of 9
Equation: Centre at the Origin
In this lesson every circle is centred at the origin $(0,0)$. The next lesson moves the centre off the origin.
Must learn
$x^{2} + y^{2} = r^{2}$
Centre $(0,0)$, radius $r$. The number on the right-hand side is $r^{2}$, never $r$.
(i) Find the centre and radius of $\,x^{2} + y^{2} = 25$
$r^{2} = 25$
$r = 5$
Centre $(0,0)$, radius $= 5$
(ii) $\,x^{2} + y^{2} = 18$
$18$ is not a perfect square — leave the radius as a surd in simplest form.
$r^{2} = 18$
$r = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$
Centre $(0,0)$, radius $= 3\sqrt{2}$
(iii) $\,x^{2} + y^{2} = 7$
$r^{2} = 7$
Centre $(0,0)$, radius $= \sqrt{7}$
(iv) $\,2x^{2} + 2y^{2} = 3$
The coefficients of $x^{2}$ and $y^{2}$ must both be $1$ before you read off $r^{2}$. Divide first.
$2x^{2} + 2y^{2} = 3$
$x^{2} + y^{2} = \dfrac{3}{2}$ (divide everything by 2)
$r^{2} = \dfrac{3}{2}$
Centre $(0,0)$, radius $= \sqrt{\dfrac{3}{2}}$
You try
State the centre and radius of $\,x^{2} + y^{2} = 49$.
$r^{2} = 49$ — take the (positive) square root.
$r^{2} = 49$
$r = 7$
Centre $(0,0)$, radius $= 7$
Centre $(0,0)$, radius $= 7$
You try
State the centre and radius of $\,3x^{2} + 3y^{2} = 12$.
Divide every term by $3$ first so the coefficients of $x^{2}$ and $y^{2}$ become $1$.
$3x^{2} + 3y^{2} = 12$
$x^{2} + y^{2} = 4$ (divide by 3)
$r^{2} = 4 \Rightarrow r = 2$
Centre $(0,0)$, radius $= 2$
Centre $(0,0)$, radius $= 2$
Section 3 of 9
Where Does the Circle Cut the Axes?
To find where a circle cuts the $x$-axis, let $y=0$. To find where it cuts the $y$-axis, let $x=0$. The same trick as for any line or curve.
Example: $\,x^{2} + y^{2} = 36$
Cuts the $x$-axis $\Rightarrow$ set $y = 0$:
$x^{2} + 0 = 36$
$x^{2} = 36 \Rightarrow x = \pm 6$
Cuts $x$-axis at $(6, 0)$ and $(-6, 0)$
Cuts the $y$-axis $\Rightarrow$ set $x = 0$:
$0 + y^{2} = 36$
Cuts $y$-axis at $(0, 6)$ and $(0, -6)$
Makes sense: for a circle at the origin with radius $r$, it must cut both axes at $\pm r$.
You try
Find all four points where $\,x^{2} + y^{2} = 25\,$ cuts the axes.
Let $y=0$ for $x$-axis cuts. Let $x=0$ for $y$-axis cuts.
$y=0$: $x^{2} = 25 \Rightarrow x = \pm 5$
$x=0$: $y^{2} = 25 \Rightarrow y = \pm 5$
$(5,0),\;(-5,0),\;(0,5),\;(0,-5)$
$(\pm 5, 0)$ and $(0, \pm 5)$
Section 4 of 9
Sketching by Guess-and-Check
When the radius is a whole number, sketching is easy — plot $(\pm r, 0)$ and $(0, \pm r)$ and join with a smooth curve. But when the radius is a surd (like $\sqrt{10}$) we still want a few integer points to plot.
The trick: guess small whole numbers for $x$ and check if $y$ comes out nicely.
Example: $\,x^{2} + y^{2} = 10$
Try $x = 1$:
$1 + y^{2} = 10 \Rightarrow y^{2} = 9 \Rightarrow y = \pm 3$
So $(1, 3)$ and $(1, -3)$ are both on the circle.
Try $x = 3$:
$9 + y^{2} = 10 \Rightarrow y^{2} = 1 \Rightarrow y = \pm 1$
So $(3, 1)$ and $(3, -1)$ are on the circle.
By symmetry $(-1, \pm 3)$ and $(-3, \pm 1)$ are too — eight clean points to plot.
Skip any $x$ where $r^{2} - x^{2}$ comes out negative (impossible) or non-square (ugly). For $x^{2}+y^{2}=10$, $x=2$ gives $y^{2}=6$ — useable as $\pm\sqrt{6}$ but not a clean integer.
You try
Find all integer points on $\,x^{2} + y^{2} = 5$.
Try $x = 0, 1, 2$. For each, work out $y^{2}$ and see if it's a perfect square.
$x=0 \Rightarrow y^{2} = 5$ (not a perfect square — skip)
$x=1 \Rightarrow y^{2} = 4 \Rightarrow y = \pm 2$ ✓
$x=2 \Rightarrow y^{2} = 1 \Rightarrow y = \pm 1$ ✓
By symmetry the negative $x$ values give the same.
$(\pm 1, \pm 2)$ and $(\pm 2, \pm 1)$ — eight integer points
$(\pm 1, \pm 2)$ and $(\pm 2, \pm 1)$
Section 5 of 9
Is the Point Inside, On, or Outside?
Sub the point's coordinates into the left-hand side of the equation and compare with $r^{2}$.
Must learn — the test
•If $\,a^{2} + b^{2} \,<\, r^{2}$ $\Rightarrow$ $(a,b)$ is inside the circle.
•If $\,a^{2} + b^{2} \,=\, r^{2}$ $\Rightarrow$ $(a,b)$ is on the circle.
•If $\,a^{2} + b^{2} \,>\, r^{2}$ $\Rightarrow$ $(a,b)$ is outside the circle.
Example: Is $(3, -5)$ inside or outside $\,x^{2} + y^{2} = 33$?
$a^{2} + b^{2} = 3^{2} + (-5)^{2} = 9 + 25 = 34$
$r^{2} = 33$
$34 > 33$
$(3,-5)$ is outside the circle
You try
Is $(2, 4)$ inside, on, or outside $\,x^{2} + y^{2} = 20$?
Work out $2^{2} + 4^{2}$ and compare with $20$.
$2^{2} + 4^{2} = 4 + 16 = 20$
$r^{2} = 20$
$20 = 20$
$(2,4)$ is on the circle
On the circle
You try
Name three points (with integer coordinates) that lie inside $\,x^{2} + y^{2} = 10$. Many answers are possible.
Any $(a,b)$ where $a^{2}+b^{2} < 10$ works. $(0,0)$ is the easiest!
$(0,0)$: $0 + 0 = 0 < 10$ ✓
$(1,2)$: $1 + 4 = 5 < 10$ ✓
$(2,2)$: $4 + 4 = 8 < 10$ ✓
e.g. $(0,0),\;(1,2),\;(2,2)$ — any with $a^{2}+b^{2} < 10$
e.g. $(0,0),\;(1,2),\;(-2,2)$ — pick any with $a^{2}+b^{2} < 10$
Section 6 of 9
Where Does a Line Meet a Circle?
Two equations, one straight, one curved. Use substitution — exactly as you did for simultaneous equations.
Method
1Make $x$ (or $y$) the subject of the line.
2Substitute into the circle equation.
3Solve the resulting quadratic for that variable.
4Sub each solution back into the line to get the matching coordinate.
(i) Line $\,x + 2y = 10\,$ and circle $\,x^{2} + y^{2} = 10$
From the line: $\;x = 10 - 2y$
Sub: $\;(10-2y)^{2} + y^{2} = 10$
$100 - 40y + 4y^{2} + y^{2} = 10$
$5y^{2} - 40y + 90 = 0$
$y^{2} - 8y + 18 = 0$ (divide by 5)
Discriminant: $b^{2} - 4ac = 64 - 72 = -8$
$-8 < 0 \;\Rightarrow\;$ no real solutions
The line does not meet the circle.
Geometrically: this line is more than $\sqrt{10}$ units away from the origin.
(ii) Line $\,2x + 3y = 5\,$ and circle $\,x^{2} + y^{2} = 2$
From the line: $\;x = \dfrac{5 - 3y}{2}$
Sub: $\;\left(\dfrac{5 - 3y}{2}\right)^{2} + y^{2} = 2$
$\dfrac{(5-3y)^{2}}{4} + y^{2} = 2$
$(5-3y)^{2} + 4y^{2} = 8$ (× 4 throughout)
$25 - 30y + 9y^{2} + 4y^{2} = 8$
$13y^{2} - 30y + 17 = 0$
$(13y - 17)(y - 1) = 0$
$y = 1$ or $y = \dfrac{17}{13}$
Back-sub into $x = \dfrac{5-3y}{2}$:
$y = 1 \Rightarrow x = \dfrac{5-3}{2} = 1$
$y = \dfrac{17}{13} \Rightarrow x = \dfrac{5 - \frac{51}{13}}{2} = \dfrac{\frac{14}{13}}{2} = \dfrac{7}{13}$
Points of intersection: $(1, 1)$ and $\left(\dfrac{7}{13}, \dfrac{17}{13}\right)$
You try
Find the points where $\,y = x + 1\,$ meets $\,x^{2} + y^{2} = 25$.
Sub $y = x+1$ into the circle. Solve for $x$, then get $y$.
$x^{2} + (x+1)^{2} = 25$
$x^{2} + x^{2} + 2x + 1 = 25$
$2x^{2} + 2x - 24 = 0$
$x^{2} + x - 12 = 0$
$(x+4)(x-3) = 0 \Rightarrow x = -4$ or $x = 3$
$x = -4 \Rightarrow y = -3$ ; $x = 3 \Rightarrow y = 4$
$(-4, -3)$ and $(3, 4)$
$(-4, -3)$ and $(3, 4)$
You try
Find the points where $\,x - y = 2\,$ meets $\,x^{2} + y^{2} = 10$.
$x = y + 2$. Sub in and solve the quadratic in $y$.
$x = y + 2$
$(y+2)^{2} + y^{2} = 10$
$y^{2} + 4y + 4 + y^{2} = 10$
$2y^{2} + 4y - 6 = 0$
$y^{2} + 2y - 3 = 0$
$(y+3)(y-1) = 0 \Rightarrow y = -3$ or $y = 1$
$y = -3 \Rightarrow x = -1$ ; $y = 1 \Rightarrow x = 3$
$(-1, -3)$ and $(3, 1)$
$(-1, -3)$ and $(3, 1)$
Section 7 of 9
Tangent at a Point on the Circle
The single most important fact for tangents: the tangent is perpendicular to the radius at the point of contact. So if we can get the slope of the radius, the tangent's slope falls out, and we have a line.
Method — always this order
1Find the slope of the radius from the centre $(0,0)$ to the point of contact $(x_{1}, y_{1})$.
2Take the perpendicular slope (flip and change sign).
3Use point-slope: $\,y - y_{1} = m(x - x_{1})\,$ with that slope and the point of contact.
Example: Tangent to $\,x^{2} + y^{2} = 13\,$ at $(-2, 3)$
First, sanity-check the point is on the circle: $(-2)^{2} + 3^{2} = 4 + 9 = 13$ ✓
Step 1. Slope of radius from $(0,0)$ to $(-2,3)$:
$m_{\text{rad}} = \dfrac{3 - 0}{-2 - 0} = -\dfrac{3}{2}$
Step 2. Perpendicular slope (flip and change sign):
$m_{\text{tan}} = \dfrac{2}{3}$
Step 3. Point-slope through $(-2, 3)$:
$y - 3 = \dfrac{2}{3}\bigl(x - (-2)\bigr)$
$3(y - 3) = 2(x + 2)$
$3y - 9 = 2x + 4$
$2x - 3y + 13 = 0$
Shortcut formula: $\,x x_{1} + y y_{1} = r^{2}$
There's a tidy formula for the tangent at $(x_{1}, y_{1})$ on $\,x^{2}+y^{2}=r^{2}\,$:
$x x_{1} + y y_{1} = r^{2}$
Quick check with our example: $x_{1}=-2$, $y_{1}=3$, $r^{2}=13$:
$-2x + 3y = 13$ $\Rightarrow$ $2x - 3y + 13 = 0$ ✓
Same answer either way — but learn the slope-of-radius method first; it generalises to any circle, including the centred-elsewhere circles in the next lesson.
You try
Find the tangent to $\,x^{2} + y^{2} = 25\,$ at $(3, -4)$. Use the slope-of-radius method.
Slope of radius $= \dfrac{-4}{3}$. Perpendicular slope $= \dfrac{3}{4}$. Now point-slope through $(3,-4)$.
Check: $9 + 16 = 25$ ✓
$m_{\text{rad}} = \dfrac{-4}{3}$
$m_{\text{tan}} = \dfrac{3}{4}$
$y - (-4) = \dfrac{3}{4}(x - 3)$
$4(y + 4) = 3(x - 3)$
$4y + 16 = 3x - 9$
$3x - 4y - 25 = 0$
$3x - 4y - 25 = 0$
You try
Find the tangent to $\,x^{2} + y^{2} = 50\,$ at $(5, 5)$. (Try the $xx_{1}+yy_{1}=r^{2}$ shortcut here.)
Plug $x_{1}=5,\,y_{1}=5,\,r^{2}=50$ straight into $xx_{1}+yy_{1}=r^{2}$.
Check: $25 + 25 = 50$ ✓
$xx_{1} + yy_{1} = r^{2}$
$5x + 5y = 50$
$x + y = 10$ (or $x + y - 10 = 0$)
$x + y - 10 = 0$
Section 8 of 9
Tangent from a Point Outside the Circle
From an external point you can usually draw two tangents to a circle. To find them, use the fact that the perpendicular distance from the centre to the tangent equals the radius.
Method
1Let the tangent be $y - y_{1} = m(x - x_{1})$ (point-slope through the external point).
2Rewrite as $\,mx - y + (y_{1} - mx_{1}) = 0\,$ (general form).
3Apply: $\;\dfrac{|\,m(0) - (0) + (y_{1} - mx_{1})\,|}{\sqrt{m^{2}+1}} = r$.
4Square both sides and solve the quadratic in $m$.
5Check if a vertical tangent $\,x = x_{1}\,$ also works (slope undefined — handled separately).
Example: Tangents from $(2, 1)$ to $\,x^{2} + y^{2} = 4$
First check: $2^{2} + 1^{2} = 5 > 4$, so $(2,1)$ is outside. Good — two tangents possible.
Let the tangent through $(2,1)$ have slope $m$:
$y - 1 = m(x - 2)$
$mx - y + (1 - 2m) = 0$
Distance from $(0,0)$ to this line $= r = 2$:
$\dfrac{|\,1 - 2m\,|}{\sqrt{m^{2}+1}} = 2$
$(1 - 2m)^{2} = 4(m^{2}+1)$
$1 - 4m + 4m^{2} = 4m^{2} + 4$
$-4m = 3$
$m = -\dfrac{3}{4}$
First tangent (sub $m = -\tfrac{3}{4}$ back):
$y - 1 = -\dfrac{3}{4}(x - 2)$
$4y - 4 = -3x + 6$
$3x + 4y - 10 = 0$
Second tangent — the vertical case. Our slope method only finds tangents with a defined slope. Check $x = 2$ separately: distance from $(0,0)$ to the line $x = 2$ is $2$, which equals the radius. So $x = 2$ is also a tangent.
Tangents: $3x + 4y - 10 = 0$ and $x = 2$
Always sanity-check for a vertical tangent. The quadratic-in-$m$ method silently drops it whenever the discriminant collapses to a single slope.
You try
Find the tangents from $(4, 3)$ to $\,x^{2} + y^{2} = 9$.
$(4,3)$ is outside: $16+9=25>9$. Let slope $=m$, line $mx-y+(3-4m)=0$, distance from origin $= 3$.
$y - 3 = m(x - 4) \Rightarrow mx - y + (3 - 4m) = 0$
$\dfrac{|3 - 4m|}{\sqrt{m^{2}+1}} = 3$
$(3-4m)^{2} = 9(m^{2}+1)$
$9 - 24m + 16m^{2} = 9m^{2} + 9$
$7m^{2} - 24m = 0$
$m(7m - 24) = 0 \Rightarrow m = 0$ or $m = \dfrac{24}{7}$
$m=0$: $y = 3$
$m = \tfrac{24}{7}$: $y - 3 = \tfrac{24}{7}(x-4) \Rightarrow 24x - 7y - 75 = 0$
$y = 3$ and $24x - 7y - 75 = 0$
$y = 3$ and $24x - 7y - 75 = 0$
You try
Find the tangents from $(5, 0)$ to $\,x^{2} + y^{2} = 9$. (Two tangents — both have a slope this time.)
$(5,0)$: $25 > 9$, outside. Line through $(5,0)$: $y = m(x-5) \Rightarrow mx - y - 5m = 0$. Distance from $(0,0)$ $= 3$.
$y = m(x-5) \Rightarrow mx - y - 5m = 0$
$\dfrac{|-5m|}{\sqrt{m^{2}+1}} = 3$
$25m^{2} = 9(m^{2}+1)$
$16m^{2} = 9 \Rightarrow m = \pm \dfrac{3}{4}$
$m = \tfrac{3}{4}$: $3x - 4y - 15 = 0$
$m = -\tfrac{3}{4}$: $3x + 4y - 15 = 0$
$3x - 4y - 15 = 0$ and $3x + 4y - 15 = 0$
$3x \pm 4y - 15 = 0$
Section 9 of 9
Finding the Equation of a Circle
With centre at $(0,0)$ fixed, you only need to find $r^{2}$. The information you're given will be one of two flavours:
Two common set-ups
AGiven a point on the circle $(x_{1}, y_{1})$ $\Rightarrow$ $r^{2} = x_{1}^{2} + y_{1}^{2}$.
BGiven a tangent line $\Rightarrow$ $r =$ perpendicular distance from $(0,0)$ to that line.
(i) Circle through $(1, -3)$ with centre $(0,0)$
$r^{2} = 1^{2} + (-3)^{2} = 1 + 9 = 10$
$x^{2} + y^{2} = 10$
(ii) Circle, centre $(0,0)$, with $\,2x + y = 5\,$ as a tangent
$r$ = perpendicular distance from $(0,0)$ to the line $2x + y - 5 = 0$:
$r = \dfrac{|\,2(0) + 1(0) - 5\,|}{\sqrt{2^{2}+1^{2}}} = \dfrac{5}{\sqrt{5}} = \sqrt{5}$
$r^{2} = 5$
$x^{2} + y^{2} = 5$
You try
Find the equation of the circle with centre $(0,0)$ that passes through $(4, -3)$.
$r^{2} = x_{1}^{2} + y_{1}^{2}$.
$r^{2} = 4^{2} + (-3)^{2} = 16 + 9 = 25$
$x^{2} + y^{2} = 25$
$x^{2} + y^{2} = 25$
You try
Find the equation of the circle with centre $(0,0)$ that has $\,3x - 4y = 20\,$ as a tangent.
$r$ = perp. distance from $(0,0)$ to $3x - 4y - 20 = 0$. Then square it.
$r = \dfrac{|3(0) - 4(0) - 20|}{\sqrt{9 + 16}} = \dfrac{20}{5} = 4$
$r^{2} = 16$
$x^{2} + y^{2} = 16$
$x^{2} + y^{2} = 16$
That's Circle 1 — centred at the origin.
Next lesson: move the centre off the origin. The form becomes $(x-h)^{2} + (y-k)^{2} = r^{2}\,$ and we'll meet the general form $\,x^{2} + y^{2} + 2gx + 2fy + c = 0$.