三点を通る円を求める

Problem

Find the equation of the circle passing through the three points $A=(0,0)$, $B=(6,0)$, and $C=(0,8)$.

Step 1: Use the general circle form

Write the circle as

$$x^2+y^2+dx+ey+f=0.$$

Since it passes through $A=(0,0)$, substitution gives $f=0$.

Step 2: Substitute the other two points

Using $B=(6,0)$:

$$36+6d=0,$$

so

$$d=-6.$$

Using $C=(0,8)$:

$$64+8e=0,$$

so

$$e=-8.$$

That leaves

$$x^2-6x+y^2-8y=0.$$

Step 3: Complete the square

Complete the square in both variables by adding $9$ and $16$:

$$x^2-6x+9+y^2-8y+16=25.$$

This becomes

$$(x-3)^2+(y-4)^2=25.$$

So the center is $(3,4)$ and the radius is $5$.

Step 4: Check with the geometric shortcut

Since $\angle A$ is a right angle, $BC$ is the diameter of the circle. With $AB=6$ and $AC=8$, the Pythagorean theorem gives

$$BC=\sqrt{36+64}=10,$$

so the radius is $5$ and the midpoint of $BC$ is $(3,4)$, matching the algebraic result.

Answer

The circle is

$$(x-3)^2+(y-4)^2=25.$$