# General Form of the Equation of a Circle

**EQUATION OF A CIRCLE: GENERAL FORM**

Expanding the standard form of the equation of the circle we derived in the last section, we’ll obtain:

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

This suggests that the most general form of the equation of a circle can be written in terms of three variables; call them *\(g\)*, *\(f\)* and *\(c\)* so that

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

Thus, the equation of the circle in terms of *\(g\)*,*\(f\)* and *\(c\)* becomes

\[\begin{align} & \boxed{{x^2} + {y^2} + 2gx + 2fy + c = 0}:\;{\text{Equation}}\;{\text{of}}\;{\text{a}}\;{\text{circle;}}\;{\text{most}}\;{\text{general}}\;{\text{form}} \hfill \\ & {\text{Centre}} = \left( { - g, - f} \right);\,{\text{Radius}} = \,\sqrt {{g^2} + {f^2} - c} \hfill \\ \end{align} \]

It should be apparent to you how the standard form and the general form of the circle’s equation are interconvertible. Which form to use where is a matter of convenience and will depend on the situation.

As a first example, let us redo Example - 1, which involves finding the equation of the circle passing through the points \((0, 0), (3, 0)\) and \((1, 2)\).

Let the equation be \(\begin{align}{x^2} + {y^2} + 2gx + 2fy + c = 0,\end{align}\) where *\(g\)*, *\(f\)* and *\(c\)* are to be determined. This equation must be satisfied by the three points through which the circle passes, and hence we’ll obtain three equations from which *\(g\)*,*\(f\)* and *\(c\)* can be determined:

\[\begin{align} {\bf{Substitute}}\;{\bf{(0,}}\;{\bf{0):}} \qquad & c = {\rm{0}}\\ {\bf{Substitute}}\;{\bf{(3,}}\;{\bf{0):}} \qquad & 9 + 6g = {\rm{0}}\\ \qquad \Rightarrow \qquad &g\,{\rm{ = }}-\frac{3}{2}\\ {\bf{Substitute}}\;{\bf{(1,}}\;{\bf{2):}} \qquad &1 + 4 - 3 + 4f = {\rm{0}}\\ \qquad \Rightarrow \qquad & f\,{\rm{ = }}-\frac{1}{2}\\ \end{align}\]

The required equation is hence:

\[{x^2} + {y^2} - 3x - y = 0\]

which is the same as what we obtained in Example - 1.

**Example - 4**

Let *C* be any circle with centre \(\begin{align}\left( {0,\sqrt 2}\right).\end{align}\) Prove that at the most two rational points can lie on *\(C\)*.

**Solution:** By a rational point, we mean a point which has both its co-ordinates rational.

Let the equation of *C* be \(\begin{align}{x^2} + {y^2} + 2gx + 2fy + c = 0\end{align}\)

We can arrive at the result easily be contradiction. Suppose that we have three rational points on the circle with the co-ordinates \(\begin{align}\left( {{x_i},\;{y_i}} \right)i = 1,\;2,\;3.\end{align}\) These three points must satisfy the equation of the circle. Thus we obtain a system of linear equations in *\(g\)* and *\(f\)*:

\[\begin{matrix} x_{1}^{2}+y_{1}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0 & \Rightarrow \\ x_{2}^{2}+y_{2}^{2}+2g{{x}_{2}}+2f{{y}_{2}}+c=0 & \Rightarrow \\ x_{3}^{2}+y_{3}^{2}+2g{{x}_{3}}+2f{{y}_{3}}+c=0 & \Rightarrow \\ \end{matrix}\left\{ \begin{matrix} \left( 2{{x}_{1}} \right)g+\left( 2{{y}_{1}} \right)f+c=-\left( x_{1}^{2}+y_{1}^{2} \right) \\ \left( 2{{x}_{2}} \right)g+\left( 2{{y}_{2}} \right)f+c=\left( -x_{2}^{2}+y_{2}^{2} \right) \\ \left( 2{{x}_{3}} \right)g+\left( 2{{y}_{3}} \right)f+c=\left( -x_{3}^{2}+y_{3}^{2} \right) \\ \end{matrix} \right\}\]

The coefficients in this system of linear equations are all rational by assumption. Thus, when we solve this system, we must obtain *\(g\)*, *\(f\)* and *\(c\)* to be all rational. But since the centre is \(\begin{align}\left( 0,\sqrt{2} \right),\end{align}\) we have \(\begin{align}f=-\sqrt{2}\end{align}\) which gives us a contradiction.

This means that our assumption of taking three rational points on the circle is wrong \(\begin{align}\Rightarrow \end{align}\) At the most two rational points can lie on this circle.

**Example - 5**

Suppose we are given two curves \(C_1\) and \(C_2\) whose equation are as follows:

\[\begin{align} {C_1}:{a_1}{x^2} + 2{h_1}xy + {b_1}{y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0\\ {C_2}:{a_2}{x^2} + 2{h_2}xy + {b_2}{y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0 \end{align}\]

It is also given that these curves intersect in four concyclic points. Prove that

\[\frac{{{a_1} - {b_1}}}{{{h_1}}} = \frac{{{a_2} - {b_2}}}{{{h_2}}}\]

**Solution:** From the discussions in the last chapter, we know that any curve \(C\) passing through the point(s) of intersection of two given curves \(\begin{align}{C_1} = 0\,\,{\rm{and}}\,\,{C_2} = 0\end{align}\) can be written as

\[\begin{align}C \equiv {C_1} + \lambda \,{C_2} = 0 \qquad \quad {\text{where}}\,\,\lambda \in \mathbb{R}\end{align}\]

We can do the same in the current example to obtain the equation of the curve passing through the four (concyclic) points of intersection as:

\[\begin{align}\left( {{a_1} + \lambda {a_2}} \right){x^2} + 2\left( {{h_1} + \lambda {h_2}} \right)xy + \left( {{b_1} + \lambda {b_2}} \right){y^2} + 2\left( {{g_1} + \lambda {g_2}} \right)x + 2\left( {{f_1} + \lambda {f_2}y} \right) + {c_1} + \lambda {c_2} = 0\end{align}\]

From the general form of the equation of the circle, we know that this equation (above) will represent the equation of a circle only if:

\[\begin{align} {\bf{Coeff}}{\bf{.}}\;{\bf{of}}\;{x^2} = {\bf{Coeff}}{\bf{.}}\;{\bf{of}}\;{y^2} \qquad & \Rightarrow \qquad {a_1} + \lambda {a_2} = {b_1} + \lambda {b_2}\\ \qquad & \Rightarrow \qquad \lambda = - \frac{{{b_1} - {a_1}}}{{{b_2} - {a_2}}} & & & ...(1)\\ {\bf{Coeff}}{\bf{.}}\;{\bf{of}}\;xy = {\bf{0}} \qquad & \Rightarrow \qquad {h_1} + \lambda {h_2} = 0\\ \qquad & \Rightarrow \qquad \lambda = - \frac{{{h_1}}}{{{h_2}}} & & & ...(2) \end{align}\]

From (1) and (2), we have

\[\begin{align}\frac{{{b_1} - {a_1}}}{{{h_1}}} = \frac{{{b_2} - {a_2}}}{{{h_2}}}\end{align}\]

**Example - 6**

Suppose that the equation of a circle is

\[\begin{align}S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0\end{align}\]

What condition must the co-ordinates of a point \(P(x_1, y_1)\) satisfy so that *\(P\)* may lie (i) inside the circle (ii) outside the circle?

**Solution:** Let the center of *\(S\)* be *\(C\)* and its radius be *\(r\)*.

The point *\(P\)* lies inside *\(S\)* if \(CP < r\) and outside *\(S\)* is \(CP > r\).

From the equation of *\(S\)*, we know *\(C\)* to be \((–g, – f)\) and *\(r\)* to be \(\begin{align}\sqrt {{g^2} + {f^2} - c} .\end{align}\) Using these facts, we can easily evaluate the required conditions:

\[\begin{align} P\;{\bf{lies}}\;{\bf{inside}}\;{\bf{the}}\;{\bf{circle:}} \qquad \qquad & C{P^2} < {r^2}\\ & {\left( {{x_1} + g} \right)^2} + {\left( {{y_1} + f} \right)^2} < {g^2} + {f^2} - c\\ & x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c < 0 \qquad \qquad \qquad ...(1)\\ P\;{\bf{lies}}\;{\bf{outside}}\;{\bf{the}}\;{\bf{circle:}} \qquad\quad & C{P^2} > {r^2}\\ & {\left( {{x_1} + g} \right)^2} + {\left( {{y_1} + f} \right)^2} > {g^2} + {f^2} - c\\ & x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c > 0 \qquad \qquad \qquad ...(2) \end{align}\]

We can write (1) and (2) concisely as

\[\boxed{\begin{align} & P\,{\text{lies inside the circle}} \qquad \Rightarrow & S\left( {{x_1},{y_1}} \right) < 0 \\ & P\,{\text{lies on the circle}} \qquad \quad \; \Rightarrow & S\left( {{x_1},{y_1}} \right) = 0 \\& P\,{\text{lies outside the circle}} \quad\; \Rightarrow & S\left( {{x_1},{y_1}} \right) > 0 \\ \end{align}} \]

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