Equation of circle .

The general standard equation for the circle centered at \(\left ( h,k \right )\) with radius \( R\) is given by $$\left ( x-h \right )^2+\left ( y-k \right )^2=R^2$$

Lets see an example - We have to write the equation of circle with center 

\(\left ( 1,2\right )\) with radius \( 4\) .

The equation will be $$\left ( x-1 \right )^2+\left ( y-2 \right )^2=4^2$$

and the figure of this equation is as shown below:

Expanded form - Expanded form of circles is simply the result of expanding the squares in standard form .

For example if we want to expand the above equation of circle i.e

 $$\left ( x-1 \right )^2+\left ( y-2 \right )^2=4^2$$

                 after expending the squares we get

$$(x^2-2x+1)+(y^2-4x+4)=16$$ 

                 or we can write it as

$$x^2+y^2-2x-4y-11=0$$

Writing the equation of circle passing through three points

Find the equation of the circle passing through the points \((2,3),\ (4,1),\ (6,5)\).
The general equation of a circle is \[ x^{2} + y^{2} + 2gx + 2fy + c = 0. \] Here \(g\), \(f\) and \(c\) are unknown constants. Step 1: Substitute the point \((2,3)\) \[ x=2,\ y=3 \] \[ 4 + 9 + 4g + 6f + c = 0 \] \[ 4g + 6f + c = -13 \tag{1} \] Step 2: Substitute the point \((4,1)\) \[ x=4,\ y=1 \] \[ 16 + 1 + 8g + 2f + c = 0 \] \[ 8g + 2f + c = -17 \tag{2} \] Step 3: Substitute the point \((6,5)\) \[ x=6,\ y=5 \] \[ 36 + 25 + 12g + 10f + c = 0 \] \[ 12g + 10f + c = -61 \tag{3} \] Now we have three linear equations: \[ \text{(1): } 4g + 6f + c = -13, \] \[ \text{(2): } 8g + 2f + c = -17, \] \[ \text{(3): } 12g + 10f + c = -61. \] Step 4: Eliminate \(c\) by subtracting equations Subtract (1) from (2): \[ (8g + 2f + c) - (4g + 6f + c)\] \[= -17 - (-13) \] \[ 4g - 4f = -4 \] \[ g - f = -1 \tag{4} \] Subtract (2) from (3): \[ (12g + 10f + c) - (8g + 2f + c)\] \[= -61 - (-17) \] \[ 4g + 8f = -44 \] \[ g + 2f = -11 \tag{5} \] Now solve equations (4) and (5). From (4): \[ g = f - 1 \] Substitute this into (5): \[ (f - 1) + 2f = -11 \] \[ 3f - 1 = -11 \] \[ 3f = -10 \] \[ f = -\frac{10}{3} \] Now substitute \(f\) into \(g = f - 1\): \[ g = -\frac{10}{3} - 1 \] \[ g = -\frac{10}{3} - \frac{3}{3} \] \[ g = -\frac{13}{3} \] Step 5: Find \(c\) using equation (1) \[ 4g + 6f + c = -13 \] \[ 4\left(-\frac{13}{3}\right) + 6\left(-\frac{10}{3}\right) + c = -13 \] \[ -\frac{52}{3} - \frac{60}{3} + c = -13 \] \[ -\frac{112}{3} + c = -13 \] Convert \(-13\) to denominator 3: \[ -13 = -\frac{39}{3} \] Thus, \[ c = -\frac{39}{3} + \frac{112}{3} \] \[ c = \frac{73}{3} \] Final Answer:} \[ \boxed{ x^{2} + y^{2} - \frac{26}{3}x - \frac{20}{3}y + \frac{73}{3} = 0 } \] This is the required equation of the circle passing through the three points.

Writing the standard form using the expanded equation of circles-  In this section we will learn how to write the given expended form of a circle back in standard form using "Complete square method"

lets see an example- we are given an equation of circle ,lets say

$$x^2+y^2+6x+4y-7=0$$

now we have to write this in standard form , we will use "completing the square method".

$$x^2+y^2+6x+4y=7$$

                                          or

$$(x^2+6x)+(y^2+4y)=7$$

Now apply the "Completing square method" we get 

$$(x^2+6x+9-9)+(y^2+4y+4-4)=7$$

$$(x^2+6x+9)+(y^2+4y+4)=7+9+4$$

$$(x+3)^2+(y+2)^2=20$$

               or we can write it as

$$(x-(-3))^2+(y-(-2))^2=20$$

                      or  we can write it as

$$(x-(-3))^2+(y-(-2))^2=(\sqrt{20})^2$$

so here we can see that the above equation is a standard form of circle, whit center  \(\left ( -3,-2 \right )\) and radius \(\sqrt{20}\). You can see the diagram of above circle.

Practice questions for circle (write following expanded forms in their standard form)

1).\(x^{2}+y^{2}+4x+4x=0\)

2).\(x^{2}+y^{2}+3x+6x=2\)

3).\(x^{2}+y^{2}+10x+10x=10\)

4).\(x^{2}+y^{2}=5\)

5).\(x^{2}+y^{2}+5x+3x=1\)