General Form of the Equation of a Circle

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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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Download SOLVED Practice Questions of General Form of the Equation of a Circle for FREE
Circles
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