Circles and triangles

Topic summary

The circumcircle of a triangle is the circle that passes through all three vertices of the triangle. The center of the circumcircle is called the circumcenter, and it is the point where the perpendicular bisectors of the sides of the triangle meet.

Is AB a Diameter for Triangle ABC?

To check if the line segment \( AB \) is the diameter of the circumcircle of triangle ABC, we use the Pythagorean theorem. If \( AB \) is the diameter, then triangle ABC must be a right-angled triangle with a right angle at vertex \( C \).

In this case, if \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) are the coordinates of the vertices of the triangle, and \( AB \) is the diameter, the condition is:

\[ AC^2 + BC^2 = AB^2 \]

To apply the Pythagorean theorem, first calculate the lengths of sides \( AC \), \( BC \), and \( AB \) using the distance formula:

For \( AC \), the distance between points \( A(x_1, y_1) \) and \( C(x_3, y_3) \) is:

\[ AC = \sqrt{(x_3 – x_1)^2 + (y_3 – y_1)^2} \]

For \( BC \), the distance between points \( B(x_2, y_2) \) and \( C(x_3, y_3) \) is:

\[ BC = \sqrt{(x_3 – x_2)^2 + (y_3 – y_2)^2} \]

For \( AB \), the distance between points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is:

\[ AB = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]

Once you have the distances \( AC \), \( BC \), and \( AB \), substitute them into the equation \( AC^2 + BC^2 = AB^2 \). If the equation holds true, then \( AB \) is the diameter of the circumcircle, and triangle ABC is a right-angled triangle with \( \angle ACB = 90^\circ \).

Finding the Circumcenter of the Triangle

The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. To find the circumcenter algebraically, we will use the following steps:

1. Find the midpoint of at least two sides of the triangle. The midpoint of a segment between points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
2. Find the gradient of each side. The gradient between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
3. Find the gradient of the perpendicular bisector. The gradient of a perpendicular line is the negative reciprocal of the original line’s gradient. If the gradient of the original line is \( m \), the gradient of the perpendicular bisector is \( -\frac{1}{m} \).
4. Write the equation of the perpendicular bisector using the point-gradient form of the line equation:
5. Set up the system of equations for the two perpendicular bisectors and solve the system to find the coordinates of the circumcenter.

For example, if the coordinates of the vertices of triangle ABC are \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \), we calculate the midpoints of sides \( AB \) and \( BC \), find the gradients of their perpendicular bisectors, and then solve the system of equations to find the circumcenter \( (h, k) \).

Finding the Equation of the Circumcircle

Once we have the coordinates of the circumcenter \( (h, k) \), we can find the equation of the circumcircle. The equation of a circle with center \( (h, k) \) and radius \( r \) is:

\[ (x – h)^2 + (y – k)^2 = r^2 \]

The radius \( r \) is the distance from the circumcenter to any of the vertices of the triangle. We can use the distance formula to calculate the radius. The distance between points \( (x_1, y_1) \) and \( (h, k) \) is:

\[ r = \sqrt{(x_1 – h)^2 + (y_1 – k)^2} \]

For example, if the circumcenter \( (h, k) \) is at \( (4, 5) \) and one of the triangle’s vertices is \( A(x_1, y_1) = (1, 2) \), then the radius is:

\[ r = \sqrt{(1 – 4)^2 + (2 – 5)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]

Thus, the equation of the circumcircle would be:

\[ (x – 4)^2 + (y – 5)^2 = (3\sqrt{2})^2 = 18 \]

Therefore, the equation of the circumcircle is:

\[ (x – 4)^2 + (y – 5)^2 = 18 \]