Co-ordinate Geometry · Paper 2
The Circle 3
Higher Level · Tangents from outside & circles from conditions · Tap NEXT to begin
Section 1 of 5
Tangents from an External Point
Identical idea to Circle 1, but the centre isn't $(0,0)$ — we measure perpendicular distance from $(-g, -f)$.
Example: Tangents from $(3, -2)$ to $\,x^{2}+y^{2}-10x-8y+21=0$
Centre and radius first:
$g = -5, \; f = -4, \; c = 21$
Centre $(5, 4), \;\; r = \sqrt{25 + 16 - 21} = \sqrt{20} = 2\sqrt{5}$
Sanity check $(3,-2)$ is outside: $(3-5)^{2}+(-2-4)^{2} = 4 + 36 = 40 > 20$ ✓
Let the tangent through $(3,-2)$ have slope $m$:
$y + 2 = m(x - 3) \;\;\Rightarrow\;\; mx - y - 3m - 2 = 0$
Distance from centre $(5,4)$ to the line $= r$:
$\dfrac{|\,5m - 4 - 3m - 2\,|}{\sqrt{m^{2}+1}} = 2\sqrt{5}$
$\dfrac{|2m - 6|}{\sqrt{m^{2}+1}} = 2\sqrt{5}$
$(2m - 6)^{2} = 20(m^{2}+1)$
$4m^{2} - 24m + 36 = 20m^{2} + 20$
$16m^{2} + 24m - 16 = 0$
$2m^{2} + 3m - 2 = 0$
$(2m - 1)(m + 2) = 0$
$m = \dfrac{1}{2}$ or $m = -2$
First tangent ($m = \tfrac{1}{2}$):
$y + 2 = \dfrac{1}{2}(x - 3) \;\Rightarrow\; x - 2y - 7 = 0$
Second tangent ($m = -2$):
$y + 2 = -2(x - 3) \;\Rightarrow\; 2x + y - 4 = 0$
Tangents: $x - 2y - 7 = 0$ and $2x + y - 4 = 0$
You try
Find the tangents from $(7, 1)$ to $\,x^{2}+y^{2}-4x-2y-4=0$.
Centre $(2,1)$, $r = 3$. Line through $(7,1)$ has equation $y - 1 = m(x - 7)$. Apply perpendicular-distance $= 3$.
$g=-2,\,f=-1,\,c=-4$: centre $(2,1)$, $r = \sqrt{4+1+4} = 3$
Line: $mx - y + (1 - 7m) = 0$
$\dfrac{|2m - 1 + 1 - 7m|}{\sqrt{m^{2}+1}} = 3 \;\Rightarrow\; \dfrac{|-5m|}{\sqrt{m^{2}+1}} = 3$
$25m^{2} = 9(m^{2}+1) \;\Rightarrow\; 16m^{2} = 9 \;\Rightarrow\; m = \pm \tfrac{3}{4}$
$m = \tfrac{3}{4}$: $3x - 4y - 17 = 0$
$m = -\tfrac{3}{4}$: $3x + 4y - 25 = 0$
$3x - 4y - 17 = 0$ and $3x + 4y - 25 = 0$
$3x - 4y - 17 = 0$ and $3x + 4y - 25 = 0$
Section 2 of 5
Circle Through Three Given Points
Three points on a circle = three equations. Three unknowns ($g$, $f$, $c$). Solve simultaneously.
Method
1Sub each point $(x, y)$ into $\,x^{2} + y^{2} + 2gx + 2fy + c = 0$.
2You'll get three linear equations in $g$, $f$, $c$.
3Subtract pairs of equations to eliminate $c$, then solve the remaining two equations for $g$ and $f$.
4Back-sub to find $c$. Read off the circle.
Example: Circle through $(5, -3)$, $(3, -1)$, $(-1, -5)$
Sub $(5, -3)$:
$25 + 9 + 10g - 6f + c = 0$
$10g - 6f + c = -34 \quad\quad (1)$
Sub $(3, -1)$:
$9 + 1 + 6g - 2f + c = 0$
$6g - 2f + c = -10 \quad\quad (2)$
Sub $(-1, -5)$:
$1 + 25 - 2g - 10f + c = 0$
$-2g - 10f + c = -26 \quad\quad (3)$
$(1) - (2)$: $4g - 4f = -24 \;\Rightarrow\; g - f = -6 \quad\quad (4)$
$(2) - (3)$: $8g + 8f = 16 \;\Rightarrow\; g + f = 2 \quad\quad (5)$
Add $(4) + (5)$: $2g = -4 \;\Rightarrow\; g = -2$
From $(5)$: $f = 2 - (-2) = 4$
Sub into $(2)$: $6(-2) - 2(4) + c = -10 \;\Rightarrow\; -20 + c = -10 \;\Rightarrow\; c = 10$
$x^{2} + y^{2} - 4x + 8y + 10 = 0$
Quick check: centre $= (2, -4)$, $r = \sqrt{4+16-10} = \sqrt{10}$. Confirm one point lies on it.
You try
Find the equation of the circle through $(0,0)$, $(6,0)$ and $(0,8)$.
Sub $(0,0) \Rightarrow c = 0$. Then sub $(6,0)$ and $(0,8)$ for $g$ and $f$.
$(0,0) \Rightarrow c = 0$
$(6,0) \Rightarrow 36 + 12g = 0 \Rightarrow g = -3$
$(0,8) \Rightarrow 64 + 16f = 0 \Rightarrow f = -4$
$x^{2} + y^{2} - 6x - 8y = 0$
$x^{2} + y^{2} - 6x - 8y = 0$
Section 3 of 5
Two Points & Centre on a Given Line
If you only know two points on the circle plus the line the centre sits on, you have three pieces of information again — enough to pin down the three unknowns $g$, $f$, $c$.
Method
1Sub each point into $x^{2} + y^{2} + 2gx + 2fy + c = 0$ $\Rightarrow$ three equations.
2Centre $(-g, -f)$ lies on the line $\Rightarrow$ sub $x = -g$, $y = -f$ into that line $\Rightarrow$ third equation.
3Solve the system.
Example: Circle through $(5, 0)$ and $(1, 2)$, centre on $\,x + 2y = -5$
Sub $(5, 0)$:
$25 + 10g + c = 0 \quad\quad (1)$
Sub $(1, 2)$:
$1 + 4 + 2g + 4f + c = 0 \Rightarrow 2g + 4f + c = -5 \quad\quad (2)$
Centre $(-g, -f)$ on $x + 2y = -5$:
$-g - 2f = -5 \Rightarrow g + 2f = 5 \quad\quad (3)$
$(1) - (2)$: $20 + 8g - 4f = 5 \Rightarrow 8g - 4f = -15 \quad\quad (4)$
From $(3)$: $g = 5 - 2f$. Sub into $(4)$:
$8(5 - 2f) - 4f = -15$
$40 - 16f - 4f = -15$
$-20f = -55 \Rightarrow f = \dfrac{11}{4}$
$g = 5 - 2 \cdot \dfrac{11}{4} = 5 - \dfrac{11}{2} = -\dfrac{1}{2}$
Sub into $(1)$:
$25 + 10 \cdot \left(-\dfrac{1}{2}\right) + c = 0 \Rightarrow 25 - 5 + c = 0 \Rightarrow c = -20$
$x^{2} + y^{2} - x + \dfrac{11}{2}y - 20 = 0$ or $2x^{2} + 2y^{2} - 2x + 11y - 40 = 0$
You try
Find the equation of the circle through $(2, 0)$ and $(0, 4)$ whose centre lies on $\,y = x + 1$.
Three equations: sub each point, plus centre $(-g,-f)$ on $y=x+1$ giving $-f = -g + 1$.
$(2,0): \;\; 4 + 4g + c = 0 \;\;(1)$
$(0,4): \;\; 16 + 8f + c = 0 \;\;(2)$
Centre on line: $-f = -g + 1 \Rightarrow g - f = 1 \;\;(3)$
$(1)-(2): \;\; -12 + 4g - 8f = 0 \Rightarrow g - 2f = 3 \;\;(4)$
$(3)-(4): \;\; f = -2 \Rightarrow g = -1$
From $(1)$: $\;\; 4 - 4 + c = 0 \Rightarrow c = 0$
$x^{2}+y^{2}-2x-4y = 0$
$x^{2}+y^{2}-2x-4y = 0$
Section 4 of 5
Circle Through a Point, with a Tangent at Another Point
Given: a circle passes through point $P$ and the line $\ell$ is tangent to it at point $T$. Find the circle.
Key idea
•Tangent at $T$ $\Rightarrow$ the radius at $T$ is perpendicular to $\ell$. The centre therefore lies on the line through $T$ perpendicular to $\ell$.
•Centre is also equidistant from $P$ and $T$ (both are on the circle).
Example: Circle through $(1, 2)$, tangent to $\,x + y = 5\,$ at $(3, 2)$
Step 1. Find the line on which the centre lies — the perpendicular to $x + y = 5$ through $(3, 2)$.
Slope of $x+y=5$ is $-1$, so perpendicular slope is $1$.
$y - 2 = 1 \cdot (x - 3) \;\Rightarrow\; y = x - 1$
So the centre is $(h, h-1)$ for some $h$.
Step 2. Centre is equidistant from $(1, 2)$ and $(3, 2)$.
$(h - 1)^{2} + (h - 1 - 2)^{2} = (h - 3)^{2} + (h - 1 - 2)^{2}$
The $(h-1-2)^{2}$ terms cancel; we're left with:
$(h - 1)^{2} = (h - 3)^{2}$
$h^{2} - 2h + 1 = h^{2} - 6h + 9$
$4h = 8 \Rightarrow h = 2$
Centre $= (2, 1)$.
Step 3. Radius = distance from centre to $(3, 2)$ (or $(1,2)$):
$r = \sqrt{(3-2)^{2} + (2-1)^{2}} = \sqrt{2}$
$(x - 2)^{2} + (y - 1)^{2} = 2$
You try
Find the equation of the circle passing through $(0, 0)$ that is tangent to the line $\,x = 4\,$ at the point $(4, 3)$.
$x = 4$ is vertical, so the perpendicular through $(4,3)$ is horizontal: $y = 3$. Centre is $(h, 3)$. Equidistant from $(0,0)$ and $(4,3)$.
Perpendicular to $x=4$ through $(4,3)$ is $y = 3$.
Centre $(h, 3)$.
Distance$^{2}$ to $(0,0)$ = Distance$^{2}$ to $(4,3)$:
$h^{2} + 9 = (h-4)^{2} + 0$
$h^{2} + 9 = h^{2} - 8h + 16$
$8h = 7 \Rightarrow h = \dfrac{7}{8}$
Centre $\left(\dfrac{7}{8}, 3\right)$, $r^{2} = \left(\dfrac{7}{8}\right)^{2} + 9 = \dfrac{49}{64} + \dfrac{576}{64} = \dfrac{625}{64}$
$\left(x - \dfrac{7}{8}\right)^{2} + (y - 3)^{2} = \dfrac{625}{64}$
$\left(x - \dfrac{7}{8}\right)^{2} + (y - 3)^{2} = \dfrac{625}{64}$
Section 5 of 5
Where a Line Meets the Circle
Substitution again — no different in principle to Circle 1, but the algebra is messier because of the extra $2gx$, $2fy$, $c$ terms.
Example: Where does $\,y = x - 1\,$ meet $\,x^{2}+y^{2}-6x-4y-3=0\,$?
$x^{2} + (x-1)^{2} - 6x - 4(x-1) - 3 = 0$
$x^{2} + x^{2} - 2x + 1 - 6x - 4x + 4 - 3 = 0$
$2x^{2} - 12x + 2 = 0$
$x^{2} - 6x + 1 = 0$
$x = \dfrac{6 \pm \sqrt{36 - 4}}{2} = \dfrac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$
Back-sub $y = x - 1$:
$y = 2 \pm 2\sqrt{2}$
$(3 + 2\sqrt{2},\;\; 2 + 2\sqrt{2})$ and $(3 - 2\sqrt{2},\;\; 2 - 2\sqrt{2})$
You try
Find where $\,y = 2x - 3\,$ meets $\,x^{2}+y^{2}-4x+2y-5=0$.
Sub $y = 2x - 3$ into the circle. Expand carefully, collect, solve the quadratic.
$x^{2} + (2x-3)^{2} - 4x + 2(2x-3) - 5 = 0$
$x^{2} + 4x^{2} - 12x + 9 - 4x + 4x - 6 - 5 = 0$
$5x^{2} - 12x - 2 = 0$
$x = \dfrac{12 \pm \sqrt{144 + 40}}{10} = \dfrac{12 \pm \sqrt{184}}{10} = \dfrac{6 \pm \sqrt{46}}{5}$
$y = 2x - 3$ for each:
$\left(\dfrac{6 + \sqrt{46}}{5},\;\dfrac{2\sqrt{46} - 3}{5}\right)$ and $\left(\dfrac{6 - \sqrt{46}}{5},\;\dfrac{-2\sqrt{46} - 3}{5}\right)$
$x = \dfrac{6 \pm \sqrt{46}}{5}$, $y = 2x - 3$ for each
That's Circle 3 — tangents from outside and circles built from conditions.
Next lesson: circles touching the axes ($c = g^{2}$ tricks), two circles touching internally or externally, and the common chord $S_{1} - S_{2} = 0$.