Standard Equation of a Circle

Let $$P\left( {x,y} \right)$$ be any point of the circle as shown in the diagram, then by the definition of a circle, the distance of point $$P\left( {x,y} \right)$$ from $$C\left( {h,k} \right)$$ must be equal to the radius of the circle $$r$$. i.e. $$\left| {CP} \right| = r$$.


As we know the distance formula from the analytic geometry as
\[\left| d \right| = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \]

Now we shall use this formula to establish the equation of the circle. Consider the points $$C\left( {h,k} \right) = \left( {{x_1},{y_1}} \right)$$ and $$P\left( {x,y} \right) = \left( {{x_2},{y_2}} \right)$$. Now using the distance formula for these two points as

\[\left| {CP} \right| = r = \sqrt {{{\left( {x – h} \right)}^2} + {{\left( {y – k} \right)}^2}} \]

Squaring both sides, we have

\[\boxed{{{\left( {x – h} \right)}^2} + {{\left( {y – k} \right)}^2} = {r^2}}\]

This is the equation of a circle with centre $$\left( {h,k} \right)$$ and radius $$r$$. This is called the equation of a circle in standard form.

Note: If the centre of the circle is at origin $$\left( {0,0} \right)$$, then $$h = 0,\,k = 0$$, so the equation of the circle takes the form

\[\boxed{{x^2} + {y^2} = {r^2}}\]

This is the equation of the circle with radius $$r$$ and the centre at the origin $$\left( {0,0} \right)$$ in two dimensions XY-plane.

Example: Find the equation of a circle with centre $$\left( {1, – 2} \right)$$ and radius $$7$$.

Solution: From the given data in the example we have centre $$\left( {1, – 2} \right)$$ and radius $$7$$. In this situation we use the standard form of equation of a circle, which is:

\[{\left( {x – h} \right)^2} + {\left( {y – k} \right)^2} = {r^2}\]

Given the condition $$h = 1,\,k = – 2$$ and $$r = 7$$, putting these values in the given equation of a circle, we have

\[\begin{gathered} {\left( {x – 1} \right)^2} + {\left( {y – \left( { – 2} \right)} \right)^2} = {\left( 7 \right)^2} \\ \Rightarrow {\left( {x – 1} \right)^2} + {\left( {y + 2} \right)^2} = 49 \\ \Rightarrow {x^2} – 2x + 1 + {y^2} + 4y + 4 = 49 \\ \Rightarrow {x^2} + {y^2} – 2x + 4y – 44 = 0 \\ \end{gathered} \]

This is the required equation of the circle.