## Diameter Form of Circle Equation

The equation of the circle drawn on the straight line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is

(\(x-x_1\))(\(x-x_2\)) + (\(y-y_1\))(\(y-y_2\)) = 0.

This is known as **diameter form of circle**.

where **\((x_1, y_1)\) and \((x_2, y_2)\)** are coordinates of two ends of diameter of circle.

**Note** – If the coordinates of the end points of a diameter of a circle are given, we can also find the equation of the circle by finding the coordinates of the center and radius. The center is the mid-point of the diameter and radius is half of the length of the diameter.

## Examples :

Example : Find the diameter form of the circle, the coordinates of the end points of whose diameter are (-1, 2) and (4, -3).

Solution : Given coordinates of two ends of diameter of circle i.e. (-1, 2) and (4, -3).

We know that the equation of the circle described on the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) as a diameter is

(\(x-x_1\))(\(x-x_2\)) + (\(y-y_1\))(\(y-y_2\)) = 0

Here, \(x_1\) = -1, \(x_2\) = 4, \(y_1\) = 2 and \(y_2\) = -3

So, the equation of the required circle is

(x + 1)(x – 4) + (y – 2)(y + 3) = 0

\(x^2 + y^2 – 3x + y – 10\) = 0

Example : Find the diameter form of the circle, drawn on the intercept made by the line 2x + 3y = 6 between the coordinates axes as diameter.

Solution : The line 2x + 3y = 6 meets X and Y axes at A(3, 0) and B(0, 2) respectively

We know that the equation of the circle described on the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) as a diameter is

(\(x-x_1\))(\(x-x_2\)) + (\(y-y_1\))(\(y-y_2\)) = 0

Taking AB as a diameter, the equation of the required circle is

(x – 3)(x – 0) + (y – 0)(y – 2) = 0

\(x^2 + y^2 – 3x – 2y\) = 0