General Equation of a Circle

Go back to  'Coordinate-Geometry'

Let us analyze in more detail the equation of a circle of radius r centered at the point \(C\left( {{x_0},{y_0}} \right)\) :

General equation of a circle

The equation of this circle is

\[{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}\]

If we square and expand, we have the following:

\[\begin{align}&{x^2} + {x_0}^2 - 2x{x_0} + {y^2} + {y_0}^2 - 2y{y_0} = {r^2}\\&\Rightarrow \,\,\,{x^2} + {y^2} - 2{x_0}x - 2{y_0}y + {x_0}^2 + {y_0}^2 - {r^2} = 0\end{align}\]

We see that there is an \({x^2}\) term, a \({y^2}\) term, a term which is linear in x, a term which is linear in y, and a constant term. Taking cue from this, we can write the general equation of a circle as follows:

\[\underbrace {{x^2} + {y^2}}_{\scriptstyle{\rm{Square \,\,terms}}\atop{\scriptstyle{\rm{with \,\,equal}}\atop\scriptstyle{\rm{coefficients}}}} + \underbrace {2gx + 2fy}_{{\rm{Linear terms}}} + \underbrace c_{\scriptstyle{\rm{Constant}}\atop\scriptstyle{\rm{term}}} = 0\]

Why have we taken the coefficients of x and y as 2g and 2f, rather than, say, g and f? These and other aspects of this general equation will become clearer when you delve deeper into this analysis at a later stage.

To summarize:

  • The equation of a circle with center \(\left( {{x_0},{y_0}} \right)\) and radius r can be written as

\[{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}\]

  • The general equation of a circle can be written as

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

If we rearrange the first equation into the general form, we have:

\[\begin{array}{l}{x^2} + {y^2} + \left( { - 2{x_0}} \right)x + \left( { - 2{y_0}} \right)y\\  \,\,\,\,\,\;\;\;\;\;\;\;\;\;\; + \left( {{x_0}^2 + {y_0}^2 - {r^2}} \right) = 0\end{array}\]

If we compare this with the general form a circle’s equation, we have:

\[\begin{align}& - 2{x_0} = 2g,\,\,\, - 2{y_0} = 2f,\,\,\,\left( {{x_0}^2 + {y_0}^2 - {r^2}} \right) = c\\&\Rightarrow \,\,\,{x_0} =  - g,\,\,\,{y_0} =  - f,\,\,\,r = \sqrt {{x_0}^2 + {y_0}^2 - c} \\&\qquad\qquad\qquad\qquad\qquad\quad\;\;= \sqrt {{g^2} + {f^2} - c} \end{align}\]

Thus, given the general form of a circle’s equation, which is

\[{x^2} + {y^2} + 2gx + 2fy + c = 0,\]

we can say that the center of this circle is

\[\left( {{x_0},{y_0}} \right) = \left( { - g, - f} \right),\]

while its radius is

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

Example 1: A circle is centered at \(\left( {3, - 2} \right)\) and has a radius of 4 units. Write its equation in the general form.

Solution: We write the equation of the circle using its center and radius, and then rearrange to the general form:

\[\begin{align}&{\left( {x - 3} \right)^2} + {\left( {y - \left( { - 2} \right)} \right)^2} = {4^2}\\&\Rightarrow \,\,\,{\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} = 16\\&\Rightarrow \,\,\,{x^2} - 6x + 9 + {y^2} + 4y + 4 = 16\\&\Rightarrow \,\,\,{x^2} + {y^2} - 6x + 4y - 3 = 0\end{align}\]

Example 2: Find the center and radius of the circle whose equation is \({x^2} + {y^2} - 4x - 6y - 12 = 0\) .

Solution: Comparing the given equation to the standard equation \({x^2} + {y^2} + 2gx + 2fy + c = 0\) , we note that

\[\begin{align}&2g =  - 4,\,\,\,2f =  - 6,\,\,\,c =  - 12\\&\Rightarrow \,\,\,g =  - 2,\,\,\,f =  - 3,\,\,\,c =  - 12\end{align}\]

Thus, the center of the circle is

\[\left( { - g, - f} \right) = \left( {2,3} \right),\]

while its radius is

\[\begin{array}{l}r = \sqrt {{g^2} + {f^2} - c} \\\,\,\, = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 3} \right)}^2} - \left( { - 12} \right)} \\\,\,\, = \sqrt {25}  = 5\end{array}\]

Example 3: (a) The equation of a circle is

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

\(P\left( {{x_1},{y_1}} \right)\) is a point not lying on this circle. What is the distance of P from the circle’s center?

(b) Find the condition so that a point \(P\left( {{x_1},{y_1}} \right)\) lies outside the circle whose equation is

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

How will this condition change if P lies inside the circle?

Solution: Observe the following figure, which shows a circle with center O and a point P not lying on the circle:

Circle and point - Distance formula

The distance OP can be calculated using the distance formula:

\[\begin{array}{l}OP = \sqrt {{{\left( {{x_1} - \left( { - g} \right)} \right)}^2} + {{\left( {{y_1} - \left( { - f} \right)} \right)}^2}} \\\,\,\,\,\,\,\,\, = \sqrt {{x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + {g^2} + {f^2}} \end{array}\]

(b) For P to lie outside the circle, the distance OP must be greater than the radius of the circle, which is \(r = \sqrt {{g^2} + {f^2} - c} \). Thus, we have:

\[\begin{array}{l}OP > r\\ \Rightarrow \,\,\,\sqrt {{x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + {g^2} + {f^2}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; > \sqrt {{g^2} + {f^2} - c} \end{array}\]

Squaring both sides, we have:

\[\begin{array}{l}{x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + {g^2} + {f^2}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; > {g^2} + {f^2} - c\\ \Rightarrow \,\,\,{x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c > 0\end{array}\]

Let us denote the circle’s equation

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

by \(S\left( {x,y} \right) = 0\) . Thus, for a point \(P\left( {{x_1},{y_1}} \right)\) to lie outside the circle \(S\left( {x,y} \right) = 0\), the required condition is

\[S\left( {{x_1},{y_1}} \right) > 0\]

Similarly, for \(P\left( {{x_1},{y_1}} \right)\) to lie inside \(S\left( {x,y} \right) = 0\) , the required condition is

\[S\left( {{x_1},{y_1}} \right) < 0\]

Of course, if \(P\left( {{x_1},{y_1}} \right)\) on the circle, then \(S\left( {{x_1},{y_1}} \right) = 0\), as the coordinates of P satisfy the circle’s equation.

Example 4: The equation of a circle is

\[{x^2} + {y^2} + 2x + 4y - 11 = 0\]

Where do the points \(P\left( {2,4} \right)\) and \(Q\left( {0,0} \right)\) lie in relation to this circle?

Solution: Represent the equation of the circle by \(S\left( {x,y} \right) = 0\). We substitute the coordinates of P and Q in the expression \(S\left( {x,y} \right)\):

\[\begin{align}&S\left( {2,4} \right)= {2^2} + {4^2} + 2\left( 2 \right) + 4\left( 4 \right) - 11\\&\qquad\quad= 4 + 16 + 4 + 16 - 11 = 29 > 0\\&S\left( {0,0} \right) = {0^2} + {0^2} + 2\left( 0 \right) + 4\left( 0 \right) - 11\\&\qquad\quad=  - 11 < 0\end{align}\]

Clearly, P lies outside the circle, while Q lies inside it.

Download practice questions along with solutions for FREE:
Coordinate Geometry
Grade 9 | Questions Set 1
Coordinate Geometry
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Coordinate Geometry
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Download practice questions along with solutions for FREE:
Coordinate Geometry
Grade 9 | Questions Set 1
Coordinate Geometry
Grade 9 | Answers Set 1
Coordinate Geometry
Grade 10 | Questions Set 1
Coordinate Geometry
Grade 10 | Answers Set 1
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