Circles with Arbitrary Centers

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Consider the following figure, which shows a circle of radius 3 units centered at the point \(C\left( { - 1,1} \right)\):

Circle with 3 units radius

What will be the equation of this circle?

Any point \(P\left( {x,y} \right)\) lying on this circle will be at a distance of 3 units from the center \(C\left( { - 1,1} \right)\) . Using the distance formula, we have:

\[\begin{align}&PC = 3\\&\Rightarrow \,\,\,\sqrt {{{\left( {x - \left( { - 1} \right)} \right)}^2} + {{\left( {y - 1} \right)}^2}}  = 3\\&\Rightarrow \,\,\,\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - 1} \right)}^2}}  = 3\\&\Rightarrow \,\,\,{\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} = 9\end{align}\]

This is the equation which the coordinates of any point on the circle will satisfy (and no point not lying on the circle will satisfy) – hence, this is the equation of the circle.

In general, consider a circle with center \(C\left( {{x_0},{y_0}} \right)\) and radius r:

Determining equation of a circle

Any point \(P\left( {x,y} \right)\) lying on this circle will be at a distance of r units from \(C\left( {{x_0},{y_0}} \right)\) . Thus,

\[\begin{align}&PC = r\\&\Rightarrow \,\,\,\sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2}}  = r\\&\Rightarrow \,\,\,{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}\end{align}\]

This is the required equation.

Example 1: Write the equation of the circle centered at the point \(\left( { - 2, - 3} \right)\) and having a radius of 4 units.

Solution: The required equation will be

\[\begin{array}{l}{\left( {x - \left( { - 2} \right)} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {4^2}\\ \Rightarrow \,\,\,{\left( {x + 2} \right)^2} + {\left( {y + 3} \right)^2} = 16\end{array}\]

Example 2: Write the equation of the circle centered at \(\left( {1, - 4} \right)\) and passing through the origin.

Solution: The following figure shows the specified circle:

Circle passing through origin

We can determine the radius r of this circle by calculating the distance between its center and the origin. We have:

\[\begin{array}{l}{r^2} = {\left( {1 - 0} \right)^2} + {\left( { - 4 - 0} \right)^2} = 17\\ \Rightarrow \,\,\,r = \sqrt {17} \end{array}\]

Thus, the equation of this circle will be

\[\begin{array}{l}{\left( {x - 1} \right)^2} + {\left( {y - \left( { - 4} \right)} \right)^2} = {\left( {\sqrt {17} } \right)^2}\\ \Rightarrow \,\,\,{\left( {x - 1} \right)^2} + {\left( {y + 4} \right)^2} = 17\end{array}\]

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Coordinate Geometry
Grade 9 | Questions Set 1
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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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