**Write equations of circles in standard form using properties :**

Here we are going to see how to write equations of circles in standard form using properties.

Usually we need two information to find the equations of circle.

(i) Center of the circle

(ii) Radius of the circle

From the given properties, first we have to find the above things and apply it in formula to get the standard equation of the circle.

Standard form of equation of circle

(x - h)^{2} + (y - k)^{2} = r^{2}

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

**Example 1 :**

Find the equation of the circle passing through the points (0, 1), (2, 3) and having the centre on the line x − 2y + 3 = 0

**Solution :**

General form of a circle :

x^{2} + y^{2} + 2gx + 2fy + c = 0

In the above equation, we may apply the given points instead of x and y.

(0, 1) lies on the circle

x = 0 and y = 1

0^{2} + 1^{2} + 2g(0) + 2f(1) + c = 0

1 + 2f + c = 0

2f + c = -1 --------(1)

(2, 3) lies on the circle

x = 2 and y = 3

2^{2} + 3^{2} + 2g(2) + 2f(3) + c = 0

4 + 9 + 4g + 6f + c = 0

4g + 6f + c = -13 --------(2)

Since the centre of the circle lies on the line x − 2y + 3 = 0, we may apply x = -g and y = -f

-g - 2(-f) + 3 = 0

-g + 2f = -3 --------(3)

By subtracting the 2nd equation from the 1st equation, we get a equation in terms of g and f

(1) - (2)

(2f + c) - (4g + 6f + c) = -1 - (-13)

2f + c - 4g - 6f - c = - 1 + 13

-4g - 4f = 12

Divide by -4, we get

g + f = -3 -------(4)

(3) + (4)

-g + 2f + (g + f) = -3 + (-3)

-g + 2f + g + f = -3 - 3

3f = -6 ==> f = -2

By applying f = -2 in 4^{th} equation, we get

g + (-2) = -3

g = -3 + 2 ==> -1

By applying f = -2 in 1^{st} equation, we get

2 (-2) + c = -1

-4 + c = -1

Add 4 on both sides

c = -1 + 4

c = 3

Hence equation of the circle is

x^{2} + y^{2} + 2(-1)x + 2(-2)y + 3 = 0

x^{2} + y^{2} - 2x - 4y + 3 = 0

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