Co-ordinate Geometry · Paper 2
The Circle 2
Higher Level · Centre $(h,k)$ & $g, f, c$ form · Tap NEXT to begin
Section 1 of 6
Centre Off the Origin
In Circle 1, every circle was centred at the origin. Now we move the centre to any point $(h, k)$.
Must learn
$(x - h)^{2} + (y - k)^{2} = r^{2}$
Centre $(h, k)$, radius $r$. Watch the signs: $(x - h)$ means centre's $x$-coordinate is $+h$, $(x + h)$ means $-h$.
(i) $(x - 3)^{2} + (y - 5)^{2} = 36$
$h = 3, \; k = 5, \; r^{2} = 36$
Centre $(3, 5)$, radius $= 6$
(ii) $(x + 2)^{2} + (y - 7)^{2} = 11$
$(x + 2)$ is the same as $(x - (-2))$ — so $h = -2$.
$h = -2, \; k = 7, \; r^{2} = 11$
Centre $(-2, 7)$, radius $= \sqrt{11}$
(iii) $x^{2} + (y + 3)^{2} = 8$
No bracket on $x$ $\Rightarrow$ $h = 0$. $(y + 3)$ $\Rightarrow$ $k = -3$.
$h = 0, \; k = -3, \; r^{2} = 8$
Centre $(0, -3)$, radius $= \sqrt{8} = 2\sqrt{2}$
You try
Find the centre and radius of $\,(x - 4)^{2} + (y + 1)^{2} = 25$.
Read off $h$, $k$, $r^{2}$. Mind the sign on $(y+1)$.
$(x-4) \Rightarrow h = 4$
$(y+1) = (y - (-1)) \Rightarrow k = -1$
$r^{2} = 25 \Rightarrow r = 5$
Centre $(4, -1)$, radius $= 5$
Centre $(4, -1)$, radius $= 5$
You try
Find the centre and radius of $\,(x + 6)^{2} + y^{2} = 20$.
No $y$-bracket $\Rightarrow k = 0$. Simplify $\sqrt{20}$.
$h = -6, \; k = 0$
$r = \sqrt{20} = 2\sqrt{5}$
Centre $(-6, 0)$, radius $= 2\sqrt{5}$
Centre $(-6, 0)$, radius $= 2\sqrt{5}$
Section 2 of 6
Building a Circle from its Pieces
If you know the centre and radius, plug straight into $(x-h)^{2} + (y-k)^{2} = r^{2}$. Some setups need an extra step to find the radius first.
(i) Centre $(3, -2)$, radius $4$
$(x - 3)^{2} + (y + 2)^{2} = 16$
(ii) Centre $(1, 5)$, passing through $(4, 9)$
Radius = distance from centre to the given point on the circle.
$r = \sqrt{(4-1)^{2} + (9-5)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
$(x - 1)^{2} + (y - 5)^{2} = 25$
(iii) Diameter with endpoints $(2, -3)$ and $(8, 5)$
The centre is the midpoint of the diameter. The radius is half the diameter (or use centre-to-endpoint).
Centre $= \left(\dfrac{2+8}{2}, \dfrac{-3+5}{2}\right) = (5, 1)$
$r = \sqrt{(8-5)^{2} + (5-1)^{2}} = \sqrt{9+16} = 5$
$(x - 5)^{2} + (y - 1)^{2} = 25$
You try
Find the equation of the circle with centre $(-2, 3)$ passing through $(1, 7)$.
$r$ = distance from $(-2,3)$ to $(1,7)$. Then plug into $(x-h)^{2}+(y-k)^{2}=r^{2}$.
$r = \sqrt{(1-(-2))^{2}+(7-3)^{2}} = \sqrt{9+16} = 5$
$r^{2} = 25$
$(x+2)^{2} + (y-3)^{2} = 25$
$(x+2)^{2} + (y-3)^{2} = 25$
You try
Find the equation of the circle with $(1,2)$ and $(7,10)$ as the endpoints of a diameter.
Centre = midpoint. Then radius = distance from centre to either endpoint.
Centre $= \left(\dfrac{1+7}{2}, \dfrac{2+10}{2}\right) = (4, 6)$
$r = \sqrt{(7-4)^{2}+(10-6)^{2}} = \sqrt{9+16} = 5$
$(x-4)^{2} + (y-6)^{2} = 25$
$(x-4)^{2} + (y-6)^{2} = 25$
Section 3 of 6
Tangent at a Point — Centre Off the Origin
Exactly the same method as Circle 1 — but the slope of the radius is now from $(h, k)$, not $(0, 0)$.
Method — unchanged from Circle 1
1Slope of radius from centre $(h, k)$ to the point of contact $(x_{1}, y_{1})$.
2Perpendicular slope (flip & change sign).
3Point-slope through the contact point.
Example: Tangent to $\,(x-3)^{2} + (y-7)^{2} = 8\,$ at $(1, 5)$
Check the point is on the circle: $(1-3)^{2} + (5-7)^{2} = 4 + 4 = 8$ ✓
Step 1. Slope of radius from $(3,7)$ to $(1,5)$:
$m_{\text{rad}} = \dfrac{5 - 7}{1 - 3} = \dfrac{-2}{-2} = 1$
Step 2. Perpendicular slope:
$m_{\text{tan}} = -1$
Step 3. Tangent through $(1, 5)$ with slope $-1$:
$y - 5 = -1(x - 1)$
$y - 5 = -x + 1$
$x + y - 6 = 0$
You try
Find the tangent to $\,(x-2)^{2} + (y+1)^{2} = 25\,$ at the point $(5, 3)$.
Centre is $(2,-1)$. Slope from $(2,-1)$ to $(5,3)$ first. Then perpendicular.
Check: $(5-2)^{2}+(3+1)^{2} = 9+16 = 25$ ✓
$m_{\text{rad}} = \dfrac{3-(-1)}{5-2} = \dfrac{4}{3}$
$m_{\text{tan}} = -\dfrac{3}{4}$
$y - 3 = -\dfrac{3}{4}(x - 5)$
$4y - 12 = -3x + 15$
$3x + 4y - 27 = 0$
$3x + 4y - 27 = 0$
Section 4 of 6
Expanding the Circle Equation
Expand $(x-h)^{2}+(y-k)^{2}=r^{2}$ and gather like terms — you'll see why exam papers usually give the equation in a "messier" form.
Example: Expand $\,(x-1)^{2} + (y+2)^{2} = 5$
$(x-1)^{2} + (y+2)^{2} = 5$
$x^{2} - 2x + 1 + y^{2} + 4y + 4 = 5$
$x^{2} + y^{2} - 2x + 4y + 5 = 5$
$x^{2} + y^{2} - 2x + 4y = 0$
Every circle's equation, no matter the centre, can be written in this expanded shape — circles of form $x^{2} + y^{2} + 2gx + 2fy + c = 0$.
Must learn — $g, f, c$ form
$x^{2} + y^{2} + 2gx + 2fy + c = 0$
•Centre: $\,(-g, -f)$
•Radius: $\,r = \sqrt{g^{2} + f^{2} - c}\,$ (must be positive under the root!)
The coefficient of $x$ is $2g$, so $g$ is half the coefficient with the same sign — then flip for the centre. Same with $f$.
Section 5 of 6
Centre & Radius from $g, f, c$
(i) $\,x^{2} + y^{2} - 6x + 4y - 12 = 0$
$2g = -6 \Rightarrow g = -3$
$2f = 4 \Rightarrow f = 2$
$c = -12$
Centre $= (-g, -f) = (3, -2)$
$r = \sqrt{g^{2}+f^{2}-c} = \sqrt{9 + 4 - (-12)} = \sqrt{25} = 5$
Centre $(3, -2)$, radius $= 5$
(ii) $\,x^{2} + y^{2} + 8x - 2y - 8 = 0$
$g = 4, \; f = -1, \; c = -8$
Centre $= (-4, 1)$
$r = \sqrt{16 + 1 + 8} = \sqrt{25} = 5$
Centre $(-4, 1)$, radius $= 5$
(iii) $\,2x^{2} + 2y^{2} - 4x + 8y - 14 = 0$
Coefficients on $x^{2}$ and $y^{2}$ must both be $1$ before reading off $g, f, c$. Divide first.
$\div 2: \;\; x^{2} + y^{2} - 2x + 4y - 7 = 0$
$g = -1, \; f = 2, \; c = -7$
Centre $= (1, -2)$
$r = \sqrt{1 + 4 + 7} = \sqrt{12} = 2\sqrt{3}$
Centre $(1, -2)$, radius $= 2\sqrt{3}$
You try
Find the centre and radius of $\,x^{2} + y^{2} - 10x + 6y + 9 = 0$.
$2g = -10$, $2f = 6$, $c = 9$. Centre is $(-g, -f)$.
$g = -5, \; f = 3, \; c = 9$
Centre $= (5, -3)$
$r = \sqrt{25 + 9 - 9} = \sqrt{25} = 5$
Centre $(5, -3)$, radius $= 5$
Centre $(5, -3)$, radius $= 5$
You try
Find the centre and radius of $\,3x^{2} + 3y^{2} + 12x - 6y - 9 = 0$.
Divide everything by $3$ first.
$\div 3: \;\; x^{2} + y^{2} + 4x - 2y - 3 = 0$
$g = 2, \; f = -1, \; c = -3$
Centre $= (-2, 1)$
$r = \sqrt{4 + 1 + 3} = \sqrt{8} = 2\sqrt{2}$
Centre $(-2, 1)$, radius $= 2\sqrt{2}$
Centre $(-2, 1)$, radius $= 2\sqrt{2}$
Section 6 of 6
Tangent at a Point — $g, f, c$ Form
Same method again. The only extra step is to find the centre first by reading off $-g$ and $-f$.
Example: Tangent to $\,x^{2}+y^{2}-4x-6y+8=0\,$ at $(1, 1)$
Find the centre:
$g = -2, \; f = -3 \Rightarrow$ centre $(2, 3)$
Check point on circle: $r^{2} = 4 + 9 - 8 = 5$. And $(1-2)^{2}+(1-3)^{2} = 1+4 = 5$ ✓
Slope of radius from $(2, 3)$ to $(1, 1)$:
$m_{\text{rad}} = \dfrac{1 - 3}{1 - 2} = \dfrac{-2}{-1} = 2$
$m_{\text{tan}} = -\dfrac{1}{2}$
$y - 1 = -\dfrac{1}{2}(x - 1)$
$2y - 2 = -x + 1$
$x + 2y - 3 = 0$
You try
Find the tangent to $\,x^{2}+y^{2}+2x-4y-20=0\,$ at the point $(3, 5)$.
Centre first ($g=1$, $f=-2$ $\Rightarrow$ $(-1,2)$). Then slope of radius from there to $(3,5)$.
$g=1, \; f=-2 \Rightarrow$ centre $(-1, 2)$
$m_{\text{rad}} = \dfrac{5-2}{3-(-1)} = \dfrac{3}{4}$
$m_{\text{tan}} = -\dfrac{4}{3}$
$y - 5 = -\dfrac{4}{3}(x - 3)$
$3y - 15 = -4x + 12$
$4x + 3y - 27 = 0$
$4x + 3y - 27 = 0$
That's Circle 2 — centre $(h,k)$ and the $g, f, c$ form.
Circle 3 picks up from here: tangents from a point outside the circle, building a circle from three points, two points with the centre on a given line, a circle through a point with a tangent at another point, and where a line meets the circle.