Equation of circle (general form)

The general form of the equation of circle is:

x² + y² + 2gx + 2fy + c = 0.

Its center and radius is given as :

Center : (-g,-f)

radius r :

$\sqrt{(g^{2} + f^{2} - c)}$

Using above expression , we can tell some more information about circle ,

If $g^{2}+f^{2}>c$ Then it is a real circle with radius r.

If $g^{2}+f^{2}= c$ Then the circle coincides with its center and it would be just a point or called point circle.

If $g^{2}+f^{2}< c$ Then this circle won’t be possible as its radius would be an imaginary number. So the solution is empty set.

Identify the center and radius of a circle with the following equation.

${\color{Red} x^{2}+y^{2} -12x-16y+19 =0 }$

First, compare the given equation with general equation of circle :

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

we get, 2g= -12

g = -6 ⇒ -g = 6

and 2f = -16

f = -8 ⇒ -f = 8

Therefore center of circle is given as : (6,8)

and radius is given as,

$r=\sqrt{g^{2}+f^{2}-c}$

$r=\sqrt{6^{2}+8^{2}-19}$

$r=\sqrt{100-19} =\sqrt{81}=9$

Find center and radius of a circle for the given equation using two ways. Firstly using general formula, secondly using standard formula by completing square method.

${\color{Red} x^{2}+4x +y^{2} -12y -41 =0 }$

Using general formula , we compare the given equation with general equation

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

we get,

2g = 4 ⇒ g= 2

and

2f = -12 ⇒ f = -6

Therefore, we get center of circle as (-g, -f) = (-2,6)

and radius as $r=\sqrt{g^{2}+f^{2}-c}$

$r=\sqrt{(-2)^{2}+6^{2}+41}$

$r=\sqrt{81}=9$

Using standard form , we complete the squares for both x and y terms.

$x^{2}+4x +y^{2} -12y -41 =0$

$x^{2}+4x +y^{2} -12y =41$

$x^{2}+4x {\color{Red} +4} +y^{2} -12y {\color{Red} +36} =41{\color{Red} +4+36}$ [ added 4 and 36 on both sides to complete perfect squares ]

$(x^{2}+4x +4) +(y^{2} -12y +36) =41 +4+36$

$(x+2)^{2} +(y-6)^{2} = 81$

Comparing it with standard form:

$\dpi{120} \left ( x-h \right )^{2}+\left ( y-k \right )^{2}=r^{2}$

we get center (h,k) = (-2,6)

and radius r²= 81 ⇒ r= 9

Stella says the equation ${\color{Red} x^{2}-8x+y^{2}+2y= 5 }$ has a center (4,-1) and radius of 5. Is she correct? why or why not ?

We can find its center using either of the above mentioned two methods.

Lets use completing square method.

$x^{2}-8x+{\color{Red} 16}+y^{2}+2y{\color{Red} +1}= 5 {\color{Red} +16+1}$ [ added 16 and 1 on both sides to complete the squares]

$(x^{2}-8x+ 16)+(y^{2}+2y +1)= 22$

$(x-4)^{2}+(y+1)^{2}= 22$

Comparing it with standard form, we get center as (4,-1)

and radius r = √22

Stella is correct that center is (4,-1) . However she didn’t find the radius correctly. Correct radius is √22 not 5.

Identify the graphs of the following equations as a circle, a point or an empty set.

a) ${\color{Red} x^{2}+y^{2}+4x=0}$

To answer this question , it would be better to use general form and above mentioned rules for radius.

Rewriting the given equation in general form we get,

$x^{2}+y^{2}+4x+0y=0$

Now comparing the given equation with general equation of circle :

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

we get,

2g = 4 ⇒ g= 2

2f = 0 ⇒ f =0

c= 0

Using formula for radius ,

$r=\sqrt{g^{2}+f^{2}-c}$

$r=\sqrt{2^{2}+0^{2}-0}= 2$

That means this is a real circle with radius 2.

b) ${\color{Red} x^{2}+y^{2}+6x-4y +15= 0 }$

Comparing the given equation with general equation of circle :

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

we get,

2g = 6 ⇒ g= 3

2f = -4 ⇒ f = -2

c= 15

Using formula for radius ,

$r=\sqrt{3^{2}+(-2)^{2}-15}$

$r=\sqrt{13-15}=\sqrt{-2}$

Since radius is an imaginary number , that means this circle not possible and its graph would be empty set.

c) ${\color{Red} x^{2}+y^{2}+6x-4y +13= 0 }$

Comparing the given equation with general equation of circle :

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

we get,

2g = 6 ⇒ g= 3

2f = -4 ⇒ f = -2

c= 13

Using formula for radius ,

$r=\sqrt{3^{2}+(-2)^{2}-13}$

$r=\sqrt{13-13}= 0$

Since radius is 0 so this would be a point .

Practice problems:

Find center and radius of a circle for the given equation using two ways. Firstly using general formula, secondly using standard formula by completing square method.

a) $x^{2}+y^{2}+2y= 5$