Co-ordinate Geometry · Paper 2
The Circle 4
Higher Level · Special cases & two circles · Tap NEXT to begin
Section 1 of 8
Circles That Touch the Axes
"Touching" an axis means the axis is a tangent to the circle. The radius drawn to the point of contact is perpendicular to the axis — so the centre is exactly $r$ units from that axis.
Must learn — derived results
•Touches the $x$-axis $\Rightarrow$ $\,|f| = r\,$ and $\,c = g^{2}$.
•Touches the $y$-axis $\Rightarrow$ $\,|g| = r\,$ and $\,c = f^{2}$.
•Touches both axes $\Rightarrow$ $\,|g| = |f| = r\,$ and $\,c = g^{2} = f^{2}$.
Where does $c = g^{2}$ come from?
Centre $(-g, -f)$, radius $\sqrt{g^{2}+f^{2}-c}$. Touches $x$-axis means the distance from the centre to $y=0$ equals $r$:
$|-f| = \sqrt{g^{2}+f^{2}-c}$
$f^{2} = g^{2} + f^{2} - c$
$c = g^{2}$
The same algebra (with $g$ and $f$ swapped) gives $c = f^{2}$ for circles touching the $y$-axis.
Example: Find both circles through $(1, -3)$ touching both axes
Both axes touched $\Rightarrow$ $|g| = |f| = r$ and $c = g^{2}$.
The point $(1, -3)$ is in the fourth quadrant. Centres of "both-axes-tangent" circles in Q4 have the form $(g_{0}, -g_{0})$ — wait, let's keep going symbolically.
Use $g = f$ or $g = -f$ (both cases). Sub $(1, -3)$ into the general form:
$1 + 9 + 2g - 6f + c = 0$
$2g - 6f + c = -10 \quad\quad (*)$
Case A: $g = f$, so $c = g^{2} = f^{2}$ ✓. Sub $f = g$, $c = g^{2}$ into $(*)$:
$2g - 6g + g^{2} = -10$
$g^{2} - 4g + 10 = 0$
Discriminant: $16 - 40 = -24 < 0$ — no real solution. Case A fails.
Case B: $g = -f$, so $c = g^{2}$ still works. Sub $f = -g$, $c = g^{2}$ into $(*)$:
$2g - 6(-g) + g^{2} = -10$
$g^{2} + 8g + 10 = 0$
$g = \dfrac{-8 \pm \sqrt{64 - 40}}{2} = \dfrac{-8 \pm \sqrt{24}}{2} = -4 \pm \sqrt{6}$
Two values of $g$ $\Rightarrow$ two circles. For each, $f = -g$ and $c = g^{2}$.
$x^{2} + y^{2} + 2(-4+\sqrt{6})x + 2(4-\sqrt{6})y + (-4+\sqrt{6})^{2} = 0$ and the partner with $\sqrt{6} \to -\sqrt{6}$.
In an exam you would normally see "nice" numbers and the discriminant comes out clean.
You try
Find the equation of a circle in the first quadrant that touches both axes and has radius $4$.
If the centre is in Q1 and the circle touches both axes, where exactly is the centre? Then write $(x-h)^{2}+(y-k)^{2}=r^{2}$.
Centre is $(4, 4)$ — distance $4$ from both axes.
$r = 4 \Rightarrow r^{2} = 16$.
$(x - 4)^{2} + (y - 4)^{2} = 16$
$(x - 4)^{2} + (y - 4)^{2} = 16$
You try
A circle touches the $y$-axis at $(0, 3)$ and passes through $(4, 3)$. Find its equation.
The radius to the contact point $(0,3)$ is perpendicular to the $y$-axis, i.e. horizontal. So the centre has $y$-coord $3$: centre $(h, 3)$. Use distance to $(4,3)$ = distance to $(0,3)$.
Centre $(h, 3)$, $r = h$ (touches $y$-axis at $x=0$).
Passes through $(4,3)$: $(h-4)^{2} + 0 = h^{2}$
$h^{2} - 8h + 16 = h^{2} \Rightarrow h = 2$
Centre $(2, 3), \;\; r = 2$.
$(x-2)^{2} + (y-3)^{2} = 4$
$(x-2)^{2} + (y-3)^{2} = 4$
Section 2 of 8
Cuts an Axis at Two Points a Given Distance Apart
If a circle cuts an axis at two points and we know the distance between them, we can pin down the radius by using the perpendicular from the centre to the chord.
Key fact
•The perpendicular from the centre to a chord bisects the chord.
Example: First-quadrant circle, tangent to $x$-axis at $(3, 0)$, cuts $y$-axis $8$ apart
Tangent to $x$-axis at $(3, 0)$ $\Rightarrow$ centre directly above, $(3, k)$ for some $k > 0$, and $r = k$.
The chord on the $y$-axis has length $8$, so half-chord $= 4$. The perpendicular distance from the centre to the $y$-axis is $3$ (the $x$-coordinate of the centre).
By Pythagoras:
$r^{2} = 3^{2} + 4^{2} = 9 + 16 = 25$
$r = 5$
So $k = r = 5$. Centre $(3, 5)$.
$(x - 3)^{2} + (y - 5)^{2} = 25$
You try
A first-quadrant circle is tangent to the $y$-axis at $(0, 4)$ and cuts the $x$-axis at two points that are $6$ units apart. Find the equation.
Tangent to $y$-axis at $(0,4) \Rightarrow$ centre is $(h, 4)$ for some $h>0$, and $r = h$. Half-chord on $x$-axis $= 3$, distance from centre to $x$-axis $= 4$. Use Pythagoras.
Centre $(h, 4)$, $r = h$.
Pythagoras: $r^{2} = 4^{2} + 3^{2} = 25 \Rightarrow r = 5$.
So $h = 5$, centre $(5, 4)$.
$(x-5)^{2} + (y-4)^{2} = 25$
$(x-5)^{2} + (y-4)^{2} = 25$
Section 3 of 8
Chord with a Given Midpoint-to-Centre Distance
Same Pythagoras idea, but now the chord might be in any direction. Given the endpoints of the chord and the distance from the centre to its midpoint, we can recover the radius and centre.
Example: Find both circles through $a(-3, 0)$ and $b(5, -4)$ with the centre $\sqrt{5}$ from the midpoint of $[ab]$
Midpoint $m$ of $[ab]$:
$m = \left(\dfrac{-3+5}{2}, \dfrac{0+(-4)}{2}\right) = (1, -2)$
Half-chord length:
$|am| = \sqrt{(1-(-3))^{2} + (-2-0)^{2}} = \sqrt{16+4} = \sqrt{20}$
Pythagoras: $r^{2} = $ (distance from centre to midpoint)$^{2}$ + (half-chord)$^{2}$:
$r^{2} = (\sqrt{5})^{2} + (\sqrt{20})^{2} = 5 + 20 = 25$
So $r = 5$. The centre lies on the perpendicular bisector of $[ab]$, at distance $\sqrt{5}$ from $m$.
Slope of $[ab]$: $\;m_{ab} = \dfrac{-4 - 0}{5 - (-3)} = -\dfrac{1}{2}$. Perpendicular slope $= 2$.
Perpendicular bisector through $m(1, -2)$:
$y + 2 = 2(x - 1) \;\Rightarrow\; y = 2x - 4$
Centre $(x, 2x-4)$ with distance $\sqrt{5}$ from $(1, -2)$:
$(x - 1)^{2} + (2x - 4 + 2)^{2} = 5$
$(x - 1)^{2} + (2x - 2)^{2} = 5$
$(x - 1)^{2} + 4(x - 1)^{2} = 5$
$5(x-1)^{2} = 5$
$(x-1)^{2} = 1 \Rightarrow x = 0$ or $x = 2$
Two centres:
$x = 0: \; (0, -4)$ ; $x = 2: \; (2, 0)$
$x^{2} + (y+4)^{2} = 25$ and $(x-2)^{2} + y^{2} = 25$
You try
Find both circles through $(1, 2)$ and $(7, 2)$ whose centre is $4$ units from the midpoint of the chord joining those two points.
Midpoint $(4, 2)$. Chord is horizontal, so perpendicular bisector is vertical: $x = 4$. Centre is $(4, 2 \pm 4)$. Use Pythagoras for $r$.
Midpoint $m = (4, 2)$.
Chord length $= 6$, half-chord $= 3$.
$r^{2} = 4^{2} + 3^{2} = 25 \Rightarrow r = 5$.
Centre on vertical line $x = 4$, distance $4$ from $m$: $(4, 6)$ or $(4, -2)$.
$(x-4)^{2} + (y-6)^{2} = 25$ and $(x-4)^{2} + (y+2)^{2} = 25$
$(x-4)^{2} + (y-6)^{2} = 25$ and $(x-4)^{2} + (y+2)^{2} = 25$
Section 4 of 8
Common Chord of Two Circles
When two circles intersect at two points, the line joining those points is their common chord.
Must learn — the trick
Common chord: $S_{1} - S_{2} = 0$
Just subtract one circle's equation from the other. The $x^{2}$ and $y^{2}$ terms cancel, leaving a linear equation — the common chord.
Example: Find the common chord of
$S_{1}: \;\; x^{2} + y^{2} + 14x - 12y + 65 = 0$
$S_{2}: \;\; x^{2} + y^{2} + 4x - 2y - 5 = 0$
$S_{1} - S_{2} = 0$:
$(14x - 4x) + (-12y - (-2y)) + (65 - (-5)) = 0$
$10x - 10y + 70 = 0$
$x - y + 7 = 0$
To find the points of intersection
Sub the common-chord equation into either circle. From $x - y + 7 = 0$: $\;y = x + 7$.
Sub into $S_{2}$:
$x^{2} + (x+7)^{2} + 4x - 2(x+7) - 5 = 0$
$x^{2} + x^{2} + 14x + 49 + 4x - 2x - 14 - 5 = 0$
$2x^{2} + 16x + 30 = 0$
$x^{2} + 8x + 15 = 0$
$(x+3)(x+5) = 0 \Rightarrow x = -3$ or $x = -5$
Back-sub $y = x + 7$:
$x = -3 \Rightarrow y = 4$ ; $x = -5 \Rightarrow y = 2$
Intersection points: $(-5, 2)$ and $(-3, 4)$
You try
Find the common chord of $\;C_{1}: \,x^{2}+y^{2}-6x+2y-15=0\;$ and $\;C_{2}: \,x^{2}+y^{2}+2x-2y-11=0$.
$C_{1} - C_{2} = 0$. The squared terms cancel — you're left with a straight-line equation.
$C_{1} - C_{2}$:
$(-6x - 2x) + (2y - (-2y)) + (-15 - (-11)) = 0$
$-8x + 4y - 4 = 0$
$2x - y + 1 = 0$
$2x - y + 1 = 0$
Section 5 of 8
Two Circles Touching
Two circles can sit relative to each other in five ways. Knowing the relationship between $d$ (the distance between centres), $r_{1}$ and $r_{2}$ tells you which.
Must learn — the five cases
1$d > r_{1} + r_{2}$ $\Rightarrow$ circles are separate (no contact)
2$d = r_{1} + r_{2}$ $\Rightarrow$ touching externally (one common point, one outside the other)
3$|r_{1} - r_{2}| < d < r_{1} + r_{2}$ $\Rightarrow$ intersecting at two points
4$d = |r_{1} - r_{2}|$ $\Rightarrow$ touching internally (one common point, one inside the other)
5$d < |r_{1} - r_{2}|$ $\Rightarrow$ one circle entirely inside the other, no contact
Example: Show $C_{1}\!: x^{2}+y^{2}+2x-2y-23=0\;$ and $\;C_{2}\!: x^{2}+y^{2}-14x-2y+41=0\;$ touch externally
Centre and radius of $C_{1}$:
$g_{1} = 1, \, f_{1} = -1, \, c_{1} = -23$
Centre $C_{1} = (-1, 1), \;\; r_{1} = \sqrt{1+1+23} = \sqrt{25} = 5$
Centre and radius of $C_{2}$:
$g_{2} = -7, \, f_{2} = -1, \, c_{2} = 41$
Centre $C_{2} = (7, 1), \;\; r_{2} = \sqrt{49+1-41} = \sqrt{9} = 3$
Distance between centres:
$d = \sqrt{(7-(-1))^{2} + (1-1)^{2}} = \sqrt{64} = 8$
Now compare:
$r_{1} + r_{2} = 5 + 3 = 8 = d$
$d = r_{1} + r_{2}\;$ — the circles touch externally.
You try
Two circles have centres $(1, 2)$ and $(5, 5)$ with radii $4$ and $1$ respectively. Do they touch? If so, how?
Compute $d$. Then check $d$ vs $r_{1}+r_{2}$ and $|r_{1}-r_{2}|$.
$d = \sqrt{(5-1)^{2}+(5-2)^{2}} = \sqrt{16+9} = 5$
$r_{1} + r_{2} = 4 + 1 = 5$
$d = r_{1} + r_{2}$
They touch externally.
Yes — touching externally ($d = 5 = r_{1}+r_{2}$).
You try
Two circles have centres $(0, 0)$ and $(3, 4)$ with radii $7$ and $2$ respectively. Describe how they sit.
$d = 5$. Check against $r_{1}+r_{2}$ and $|r_{1}-r_{2}|$.
$d = \sqrt{9 + 16} = 5$
$r_{1} + r_{2} = 9$, $|r_{1} - r_{2}| = 5$
$d = |r_{1} - r_{2}|$
Touching internally — small circle inside the big one, sharing one point.
Touching internally ($d = 5 = |r_{1}-r_{2}|$).
Section 6 of 8
Find an Unknown Radius from a Touching Condition
If you know two circles touch (externally or internally) and you know one of them fully but only the centre of the other, you can recover the missing radius using $\,d = r_{1} + r_{2}\,$ or $\,d = |r_{1} - r_{2}|$.
Example: Circle $S$ has centre $(8, 5)$, radius $6$. Circle $K$ has centre $(2, -3)$. $S$ lies inside $K$ and they touch. Find $K$'s radius.
$S$ inside $K$, touching $\Rightarrow$ internal touch, with $K$ the larger circle:
$d = r_{K} - r_{S}$
Compute $d$:
$d = \sqrt{(2-8)^{2}+(-3-5)^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$
$10 = r_{K} - 6$
$r_{K} = 16$
And the equation of $K$:
$(x - 2)^{2} + (y + 3)^{2} = 256$
You try
Circle $A$ has centre $(0, 0)$ and radius $3$. Circle $B$ has centre $(7, 0)$ and touches $A$ externally. Find $B$'s radius.
$d = 7$. External touch $\Rightarrow d = r_{A} + r_{B}$.
$d = 7$
$r_{A} + r_{B} = d$
$3 + r_{B} = 7$
$r_{B} = 4$
$r_{B} = 4$
You try
A small circle has centre $(0, 0)$ and radius $2$. A large circle, centre $(0, 3)$, contains it and they touch internally. Find the radius of the large circle.
$d = 3$. Internal touch with the small circle inside the big one: $d = R - r$ where $R$ is the big radius.
$d = 3$
$d = R - r \Rightarrow 3 = R - 2$
$R = 5$
$R = 5$
Section 7 of 8
Common Tangent at the Point of Contact
When two circles touch (either externally or internally), they share a single common tangent at the point of contact — and the $S_{1} - S_{2} = 0$ trick gives it for free, just like the common chord.
Key fact
•If $S_{1}$ and $S_{2}$ touch (intersect at exactly one point), then $\,S_{1} - S_{2} = 0\,$ is the equation of their common tangent at that point.
•This is exactly the same calculation as the common chord — the geometry just collapses to a single point of contact instead of two.
Example: Common tangent of the touching circles from Section 5
$C_{1}: x^{2}+y^{2}+2x-2y-23=0$ and $C_{2}: x^{2}+y^{2}-14x-2y+41=0$ touch externally.
$C_{1} - C_{2} = 0$:
$(2x - (-14x)) + (-2y - (-2y)) + (-23 - 41) = 0$
$16x + 0 - 64 = 0$
$x = 4$ (the common tangent — a vertical line)
Sanity check: distance from $C_{1}$'s centre $(-1, 1)$ to $x = 4$ is $5 = r_{1}$ ✓, and distance from $C_{2}$'s centre $(7, 1)$ to $x = 4$ is $3 = r_{2}$ ✓. Both circles touch the line — it really is the common tangent.
You try
Two circles touch externally: $C_{1}: x^{2}+y^{2}=4$ and $C_{2}: x^{2}+y^{2}-10x+21=0$. Find the equation of the common tangent at the point of contact.
Rewrite $C_{1}$ as $x^{2}+y^{2}-4=0$. Then do $C_{1}-C_{2}=0$.
$C_{1}: x^{2}+y^{2}-4=0$
$C_{2}: x^{2}+y^{2}-10x+21=0$
$C_{1} - C_{2}: \;\; 10x - 25 = 0$
$x = \dfrac{5}{2}$
$x = \dfrac{5}{2}$
Section 8 of 8
Putting it all Together — Exam-Style
Real exam questions stitch several of these ideas into one. Here are two compounds — pause before each step to plan, then execute.
Compound problem 1
A circle has centre $(3, 2)$ and passes through $(6, 6)$. Find its equation. Then find the equation of the tangent at $(6, 6)$.
Radius:
$r = \sqrt{(6-3)^{2}+(6-2)^{2}} = \sqrt{9+16} = 5$
Circle: $(x-3)^{2} + (y-2)^{2} = 25$
Tangent at $(6,6)$:
$m_{\text{rad}} = \dfrac{6-2}{6-3} = \dfrac{4}{3}$ $\Rightarrow$ $m_{\text{tan}} = -\dfrac{3}{4}$
$y - 6 = -\dfrac{3}{4}(x - 6)$
$4y - 24 = -3x + 18$
Tangent: $3x + 4y - 42 = 0$
Compound problem 2
$C_{1}$ has equation $\,x^{2}+y^{2}-4x-2y-20=0$. $C_{2}$ has centre $(5, 4)$ and touches $C_{1}$ externally. Find $C_{2}$'s radius, then the equation of $C_{2}$, then the common tangent.
$C_{1}$ in a useful form:
$g_{1} = -2, \, f_{1} = -1, \, c_{1} = -20$
Centre $C_{1} = (2, 1), \;\; r_{1} = \sqrt{4+1+20} = 5$
Distance between centres:
$d = \sqrt{(5-2)^{2}+(4-1)^{2}} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$
External touch: $d = r_{1} + r_{2}$
$3\sqrt{2} = 5 + r_{2} \Rightarrow r_{2} = 3\sqrt{2} - 5$
Numerically $\approx 4.24 - 5 < 0$ — so the configuration described is impossible. (In an exam, this would be your cue to re-read the question.) For a clean version with positive radius, see the You-Try below.
You try (compound)
$C_{1}: x^{2}+y^{2}-4x-2y-4=0$. $C_{2}$ has centre $(8, 5)$ and touches $C_{1}$ externally. Find $r_{2}$ and the equation of $C_{2}$.
Centre and radius of $C_{1}$ first. Then $d$ between centres. Then $d = r_{1} + r_{2}$.
$C_{1}$: centre $(2, 1)$, $r_{1} = \sqrt{4+1+4} = 3$
$d = \sqrt{(8-2)^{2}+(5-1)^{2}} = \sqrt{36+16} = \sqrt{52}$
Hmm, $\sqrt{52}$ isn't clean. $d \approx 7.21$, so $r_{2} \approx 4.21$.
$r_{2} = \sqrt{52} - 3 = 2\sqrt{13} - 3$
$r_{2} = 2\sqrt{13} - 3$, $C_{2}: (x-8)^{2}+(y-5)^{2} = (2\sqrt{13}-3)^{2}$
$r_{2} = 2\sqrt{13} - 3$, $(x-8)^{2}+(y-5)^{2} = (2\sqrt{13}-3)^{2}$
You try (final)
$C_{1}$ has centre $(1, 1)$ and radius $2$. $C_{2}$ has centre $(7, 9)$ and touches $C_{1}$ externally. Find $r_{2}$ and the common tangent at the point of contact.
$d = \sqrt{36+64} = 10$. So $r_{2} = 10 - 2 = 8$. Then $S_{1} - S_{2} = 0$ for the common tangent.
$d = \sqrt{(7-1)^{2}+(9-1)^{2}} = \sqrt{100} = 10$
$d = r_{1} + r_{2} \Rightarrow r_{2} = 10 - 2 = 8$
$C_{1}: (x-1)^{2}+(y-1)^{2}=4 \Rightarrow x^{2}+y^{2}-2x-2y-2=0$
$C_{2}: (x-7)^{2}+(y-9)^{2}=64 \Rightarrow x^{2}+y^{2}-14x-18y+66=0$
$C_{1}-C_{2}$: $12x + 16y - 68 = 0$
$r_{2} = 8$ ; common tangent: $3x + 4y - 17 = 0$
$r_{2} = 8$, common tangent $3x + 4y - 17 = 0$
That's the Circle topic done.
You've covered: centred at origin, centred at $(h,k)$, general form, finding circles from points or tangents, tangents at points and from external points, axis-touching cases, two circles touching, common chords and common tangents. Practice past papers next — they reward the slope-of-radius and $S_{1}-S_{2}=0$ tricks above all.