CirclesEdexcel A-Level Maths: Revision notes

A circle with centre $(a,b)$ and radius $r$ has equation $(x-a)^2+(y-b)^2=r^2.$ If the centre is the origin this becomes $x^2+y^2=r^2$. Every point on the circle is a distance $r$ from the centre, so the equation is just Pythagoras applied to the distance between $(x,y)$ and $(a,b)$. To test a point, find its distance squared from the centre and compare it with $r^2$: smaller means inside, equal means on the circle, larger means outside. For the circle $(x-3)^2+(y+2)^2=25$ the centre is $(3,-2)$ and $r=5$. The point $(6,3)$ gives $3^2+5^2=34>25$, so it is outside. If the endpoints of a diameter are given, the centre is their midpoint and the radius is half the distance between them.

Expanding $(x-a)^2+(y-b)^2=r^2$ gives the general form $x^2+y^2+2fx+2gy+c=0,$ with centre $(-f,-g)$ and $r^2=f^2+g^2-c$. The circle exists only if $r^2>0$. In practice you complete the square in $x$ and in $y$ rather than memorise this. For $x^2+y^2-6x+4y-12=0$: $(x-3)^2-9+(y+2)^2-4-12=0\ \Rightarrow\ (x-3)^2+(y+2)^2=25.$ The centre is $(3,-2)$ and the radius is $5$. To go the other way, expand the brackets and collect all terms on one side. The coefficients of $x^2$ and $y^2$ must be equal (divide through first if they are both $2$, say).

Three properties are used constantly in coordinate geometry:

• Angle in a semicircle is a right angle. If $AB$ is a diameter and $P$ is on the circle, then $AP\perp BP$, so the product of their gradients is $-1$.
• The perpendicular from the centre to a chord bisects the chord. So the perpendicular bisector of any chord passes through the centre.
• The radius at a point is perpendicular to the tangent at that point. Perpendicular lines have gradients $m_1$ and $m_2$ with $m_1m_2=-1$. The converse of the first property also helps: if angle $ADB=90^\circ$ then $D$ lies on the circle with diameter $AB$.

To find the tangent at a point $A$ on a circle with centre $C$:

1. Find the gradient of the radius $CA$.
2. The tangent gradient is the negative reciprocal.
3. Use $y-y_1=m(x-x_1)$ with the coordinates of $A$. Example: $C(2,1)$ and $A(5,5)$. The radius gradient is $\frac{4}{3}$, so the tangent gradient is $-\frac34$ and $y-5=-\frac34(x-5)$, giving $3x+4y-35=0$. The normal at $A$ is perpendicular to the tangent, so it is the line $CA$ itself and passes through the centre. A tangent and its radius also form a right angle, which gives lengths and areas by Pythagoras.

The circumcircle passes through all three vertices. Its centre is equidistant from the vertices, so it lies on the perpendicular bisector of each side. Method for vertices $P$, $Q$, $R$:

1. Find the midpoint and gradient of $PQ$, then the equation of its perpendicular bisector.
2. Do the same for $QR$.
3. Solve the two equations simultaneously for the centre.
4. Find the radius as the distance from the centre to any vertex.
5. Write $(x-a)^2+(y-b)^2=r^2$. For $P(0,6)$, $Q(7,5)$, $R(6,-2)$ the bisectors are $y=7x-19$ and $x+7y=17$, meeting at $(3,2)$, and $r=5$. If a triangle has a right angle, the hypotenuse is a diameter of the circumcircle, so the centre is the midpoint of the hypotenuse. Checking the distance to the third vertex is a good way to catch arithmetic slips.