Find circumcenter of triangle with vertices (0,0) (4,0) (0,3)

Question

Find the circumcenter of a triangle with vertices $A(0, 0)$ , $B(4, 0)$ , and $C(0, 3)$ .

Solution — Step by Step

The circumcenter is the point equidistant from all three vertices of a triangle. It is the center of the circumscribed circle (the circle that passes through all three vertices). We find it by solving the system: $PA = PB = PC$ , where $P(x, y)$ is the circumcenter.

Let the circumcenter be $P(x, y)$ .

Condition 1: $PA = PB$

\sqrt{x^2 + y^2} = \sqrt{(x-4)^2 + y^2}

Squaring both sides:

x^2 + y^2 = (x-4)^2 + y^2

x^2 = x^2 - 8x + 16

8x = 16

x = 2

Condition 2: $PA = PC$

\sqrt{x^2 + y^2} = \sqrt{x^2 + (y-3)^2}

Squaring both sides:

x^2 + y^2 = x^2 + y^2 - 6y + 9

6y = 9

y = \frac{3}{2} = 1.5

The circumcenter is $P\!\left(2,\ \dfrac{3}{2}\right)$ .

We can verify:

$PA = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$
$PB = \sqrt{(2-4)^2 + \frac{9}{4}} = \sqrt{4 + \frac{9}{4}} = \frac{5}{2}$
$PC = \sqrt{4 + (1.5-3)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = \frac{5}{2}$

All three distances equal $\dfrac{5}{2}$ — confirmed.

Circumcenter = $\left(2,\ \dfrac{3}{2}\right)$ , Circumradius = $\dfrac{5}{2}$

Why This Works

The circumcenter is the intersection of the perpendicular bisectors of the three sides. When we set $PA = PB$ , we are finding the locus of points equidistant from A and B — which is exactly the perpendicular bisector of AB. The intersection of two such loci gives the unique point equidistant from all three vertices.

For a right-angled triangle, the circumcenter always lies at the midpoint of the hypotenuse. Let’s verify: the hypotenuse is BC (from $(4,0)$ to $(0,3)$ ), and its midpoint is $\left(\dfrac{4+0}{2}, \dfrac{0+3}{2}\right) = \left(2, \dfrac{3}{2}\right)$ . This matches our answer.

Alternative Method

Using perpendicular bisectors:

Midpoint of AB = $(2, 0)$ . AB is along the x-axis, so its perpendicular bisector is $x = 2$ .

Midpoint of AC = $(0, 1.5)$ . AC is along the y-axis, so its perpendicular bisector is $y = 1.5$ .

Intersection: $x = 2$ , $y = 1.5$ → circumcenter = $\left(2, \dfrac{3}{2}\right)$ . Same answer, and much faster for right-angled triangles!

For any right-angled triangle, the circumcenter is the midpoint of the hypotenuse. This shortcut works here because angle A = 90° (vertex at origin where the two legs meet along the axes). In competitive exams, identify the right angle first — it immediately tells you where the circumcenter is.

Common Mistake

Many students confuse circumcenter with centroid. The centroid is $\left(\dfrac{0+4+0}{3}, \dfrac{0+0+3}{3}\right) = \left(\dfrac{4}{3}, 1\right)$ , which is different from the circumcenter $\left(2, \dfrac{3}{2}\right)$ . The centroid is the “centre of mass” of the triangle; the circumcenter is the centre of the circumscribed circle. Always re-read the question to confirm which you need.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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