The points \((3,8)\) and \((5,2)\) lie at the ends of a diameter of a circle. What is the equation of the circle?
The equation of a circle can always be written as \[(x-a)^2+(y-b)^2=r^2,\] where the centre is \((a,b)\) and the radius is \(r\). So if we can find the centre and the radius, we can just write down the equation.
The centre of the circle is the midpoint of a diameter, so the centre is at \((\frac{3+5}{2},\frac{8+2}{2})=(4,5)\).
The radius is the distance from the centre to any point on the circumference, so using Pythagoras, it is \[r=\sqrt{(4-3)^2+(5-8)^2}=\sqrt{1^2+(-3)^2}=\sqrt{10}.\] As a check, if we use the other point on the circumference, we get \[r=\sqrt{(4-5)^2+(5-2)^2}=\sqrt{(-1)^2+3^2}=\sqrt{10}.\]
Alternatively, we could find the (length of the) diameter and halve it to get the radius: the diameter is \[d=\sqrt{(3-5)^2+(8-2)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10},\] so \(r=\sqrt{10}\).
Therefore the circle has equation \[(x-4)^2+(y-5)^2=10\] using the general formula \((x-a)^2+(y-b)^2=r^2\) for a circle with centre \((a,b)\) and radius \(r\).
In fact, we can save ourselves some effort by realising that we don’t need to take square roots in this problem: we only need \(r^2\) for the formula.
So we could have used Pythagoras to write \[r^2=(4-3)^2+(5-8)^2=1^2+(-3)^2=10\] instead, and then the circle’s equation is just \((x-4)^2+(y-5)^2=10\).
If you use this approach, you just have to be careful to remember that you already have \(r^2\) and that you don’t need to square it again. Writing \((x-4)^2+(y-5)^2=100\) would be unfortunate!
Some more points…
In the previous question, two points were enough to specify a circle. But is that always the case? For each of the following three statements, is the answer ALWAYS, SOMETIMES or NEVER? Do justify your answers!
If we specify two (distinct) points in the plane, there is a unique circle that passes through them?
If we specify three (distinct) points in the plane, there is a unique circle that passes through them?
If we specify four (distinct) points in the plane, there is a unique circle that passes through them?
Never true: there are infinitely many circles passing through any two distinct points. All of their centres lie on a straight line, which is the perpendicular bisector of the line joining the two given points.
Sometimes true: If the three points lie on a straight line, then there is no circle through them. In all other cases, there is a unique circle through the three points.
Sometimes true: Three of the points (as long as they are not on a straight line) will have a unique circle through them. The fourth point might lie on that circle or might not. Points which all lie on a single circle are called concyclic. (You might also recall coming across cyclic quadrilaterals; a quadrilateral is cyclic if all of its vertices lie on the circumference of a circle.)
