Circle properties and tangentsAQA A-Level Maths: Revision notes

The angle in a semicircle is a right angle: if $AB$ is a diameter and $P$ is any other point on the circle, then $A\hat{P}B=90^\circ$. The converse also holds: if $A\hat{P}B=90^\circ$ then $P$ lies on the circle with diameter $AB$. In coordinates, find the circle with diameter $AB$ (centre at the midpoint, $r=\frac12AB$) and test whether $P$ is on it. For $A(-1,1)$ and $B(7,5)$ the centre is $(3,3)$ and $r^2=20$. $P(7,1)$ gives $16+4=20$, so $A\hat{P}B=90^\circ$. A point inside the circle gives an obtuse angle and a point outside gives an acute angle.

A chord is a straight line joining two points on a circle. The perpendicular from the centre to a chord bisects the chord, so it meets the chord at its midpoint $M$. Conversely, the line from the centre to the midpoint of a chord is perpendicular to the chord, and the perpendicular bisector of any chord passes through the centre. This makes a right-angled triangle $OMP$ with $OP=r$, $MP=\frac12PQ$: $r^2=OM^2+MP^2$. For $O(2,3)$ and chord $P(-1,7)$, $Q(6,0)$: $M=\left(\frac52,\frac72\right)$, $OM^2=\frac12$, $MP^2=\frac{49}{2}$, and $OM^2+MP^2=25=r^2$.

A tangent touches the circle at exactly one point. The radius at that point is perpendicular to the tangent. So if the tangent point is $T$ and the centre is $K$, the tangent gradient is $-\frac{1}{m_{KT}}$. This means the tangent and radius gradients multiply to $-1$ (unless one is horizontal and the other vertical).

Method: (1) find the centre $K$; (2) find the gradient of the radius $KT$; (3) take the negative reciprocal; (4) use $y-y_1=m(x-x_1)$ with the point $T$. Example: $(x-1)^2+(y+2)^2=40$ at $T(7,-4)$: $K(1,-2)$, $m_{KT}=\frac{-4+2}{7-1}=-\frac13$, tangent gradient $3$, so $y+4=3(x-7)$, i.e. $3x-y-25=0$. Check that $T$ lies on the circle first if it is not given: $36+4=40$.

Because the radius is perpendicular to the tangent, triangle $KTP$ (with $P$ on the tangent) has a right angle at $T$, so $PT^2=KP^2-r^2$. Example: the tangent $3x-y-25=0$ meets the $x$-axis at $P\left(\frac{25}{3},0\right)$; $KP^2=\frac{520}{9}$ and $r^2=40$, so $PT^2=\frac{160}{9}$ and $PT=\frac{4\sqrt{10}}{3}$. Questions often combine the three properties: use the semicircle angle for a right angle at the circumference, the chord property for a midpoint, and the tangent property for a right angle at the point of contact.