Introduction To Equations Of Circles

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As we know very well from pure geometry, a circle is a geometrical figure described by a moving point in the Euclidean plane such that its distance from a fixed point is always constant. The fixed point is called the center of the circle while the fixed distance is called the radius of the circle.

Moving over to a co-ordinate system, let us denote the center \(C \) of the circle by \(({x_0}, {y_0})\) and the radius by \(r\). For any point \(P(x , y)\) lying on the circle, the length \(PC\) must be equal to \(r\). Using the distance formula for \(PC , \) we therefore obtain:

\[\boxed{{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} = {r^2}}:{\text{Equation}}\,{\text{of}}\,{\text{the}}\,{\text{circle}}\]

This equation must be satisfied by every point\( P(x , y)\) lying on the circle; therefore, this is the equation that uniquely describes the given circle. We simply call it the equation of the circle, with center \((x_0 , y_0 )\) and   radius \(r\).

For example, consider the circle with its center at \((1, 1)\) and radius equal to unity:

The equation of the this circle is

\[\begin{align}   {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} &= {1^2} \hfill \\    \Rightarrow \qquad {x^2} + {y^2} - 2x - 2y + 1 &= 0 \hfill \\  \end{align} \]

Example - 1

Find the equation of the circle passing through the points \((0, 0), (3, 0) \)and \((1, 2)\).

Solution: To write the equation of the required circle, we must find its center and radius.

Recall from pure geometry that a circle can always be drawn through three non-collinear points. This can be done as follows: join the points to form a triangle. Draw the perpendicular bisectors of any of the two sides of this triangle. Their point of intersection gives us the center\( C\). The distance of \( C\) from any of the vertices gives the radius \(r\) of the circles:

We apply this result to the current example:

 

The point \( C\) is the intersection of the two angle bisectors

\[\begin{align}\Rightarrow \qquad & C \equiv \left( {\frac{3}{2},\frac{1}{2}} \right)\end{align}\]

We can now easily evaluate the radius r as the length \(OC\):

\[\begin{align}r = OC = \sqrt {\frac{9}{4} + \frac{1}{4}}  = \frac{{\sqrt {10} }}{2}\end{align}\]

Finally, the equation of the required circle becomes:

\[\begin{align}{\left( {x - \frac{3}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} = \frac{{10}}{4}\end{align}\]

We will subsequently see another method to solve this type of questions.

Example - 2

Find the equation of the circle which touches the co-ordinate axes and whose centre lies on the \(\begin{align}x - 2y = 3\end{align}\) line

Solution: A circle of radius r touching the co-ordinate axes can be in one of the four following configurations, with four corresponding equations mentioned alongside:

Note from these four possible cases that the centre of such a circle either lies on \(y = x \)or on \(y = - x \).

In the current example, the centre is also given to lie on \(x – 2y = 3\). Thus, there will be two circles, with the two centres being given by the point of intersection \(y = x\) and \(y = - x \) with \(x – 2y = 3.\)

\[\begin{align}  y = x\,{\rm{and}}\,\,x - 2y = 3 &  \Rightarrow  \qquad {C_1} \equiv \left( { - 3, - 3} \right)\\
 &\Rightarrow  \qquad {\rm{Equation}}\;{\rm{of}}\;{\rm{the}}\;{\rm{circle}}\;{\rm{is}}\;{\left( {x + 3} \right)^2} + {\left( {y - 3} \right)^2} = 9\\  y =  - x\,{\rm{and}}\,\,x - 2y = 3 &  \Rightarrow  \qquad {C_2} \equiv \left( {1,\, - 1} \right)\\  & \Rightarrow  \qquad {\rm{Equation}}\;{\rm{of}}\;{\rm{the}}\;{\rm{circle}}\;{\rm{is}}\;{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 1  \end{align}\]

Example - 3

Find the equation of the circle with radius 5 and which touches another circle \(\begin{align}{x^2} + {y^2} - 2x - 4y - 20 = 0\end{align}\) externally at the point \((5, 5)\)

Solution: Let us first try to rearrange the equation of the given circle in the standard form from which we’ll be able to deduce its centre and radius:

\[\begin{align}  & {x^2} + {y^2} - 2x - 4y - 20 = 0\\   \Rightarrow \qquad & {\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = 25  \end{align}\]

Therefore, the centre of this circle is \((1, 2)\) and its radius is \(5\).

We should now draw a geometrical figure which will certainly make things more clear:

As explained in the figure, we can now evaluate \(\begin{align}({x_0},\;{y_0})\end{align}\), the centre of the required circle:

\[\begin{align}\frac{{{x_0} + 1}}{2} = 5\,\, \Rightarrow \,\,{x_0} = 9 \qquad \qquad \qquad  \frac{{{y_0} + 2}}{2} = 5\,\, \Rightarrow \,\,{y_0} = 8\end{align}\]

Thus, the required equation is

\[{\left( {x - 9} \right)^2} + {\left( {y - 8} \right)^2} = 25\]

Download SOLVED Practice Questions of Introduction To Equations Of Circles for FREE
Circles
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Download SOLVED Practice Questions of Introduction To Equations Of Circles for FREE
Circles
grade 11 | Questions Set 1
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grade 11 | Questions Set 2
Circles
grade 11 | Answers Set 2
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