**Equation of circle with extremities of diameter are given : **

Here we are going to see how to find the equation of circle with extremities of diameter are given.

Equation of circle when endpoints of the diameter are given :

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

Here (x_{1}, y_{1}) and (x_{2}, y_{2}) are the endpoints of the circle.

Let us look into some example problems to understand the above concept.

**Example 1 :**

Find the equation of the circle if (2, − 3) and (3, 1) are the extremities of a diameter.

**Solution :**

Here (x_{1}, y_{1}) ==> (2, -3) and (x_{2}, y_{2}) ==> (3, 1)

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

(x - 2) (x - 3) + (y - (-3)) (y - 1) = 0

(x - 2) (x - 3) + (y + 3) (y - 1) = 0

x^{2} - 3x - 2x + 6 + y^{2} - y + 3y - 3 = 0

x^{2} - 5x + y^{2} + 2y + 6 - 3 = 0

x^{2} + y^{2} - 5x + 2y + 3 = 0

**Example 2 :**

Find the equation of the circle described on the line joining the points (1, 2) and (2, 4) as its diameter.

**Solution :**

Here (x_{1}, y_{1}) ==> (1, 2) and (x_{2}, y_{2}) ==> (2, 4)

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

(x - 1) (x - 2) + (y - 2) (y - 4) = 0

x^{2} - 2x - x + 2 + y^{2} - 4y - 2y + 8 = 0

x^{2} - 3x + 2 + y^{2} - 6y + 8 = 0

x^{2} + y^{2} - 3x - 2y + 2 + 8 = 0

x^{2} + y^{2} - 3x - 2y + 10 = 0

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