Just as we model a line by a given linear equation, we need an equation to model the properties of a circle. Indeed, an equation is what defines each curve and its properties. In a similar way, we will here develop the equation of a circle which will help to model its properties on a cartesian plane.
Equation of a Circle with centre and radius (standard form)
Borrowing from the definition of a circle, recall that
A circle is the set of all the points that are equidistant from a given fixed point.
Translating the definition into an equation, we get
\[OP^2=(x-h)^2+(y-k)^2\]
where \((x,y)\) represents all the points on the circle and, hence, it varies. is the fixed point from which the distance is measured. The coordinates of the fixed point mentioned earlier are of the Centre of the circle from which the distance to all the points is measured. The coordinates are the variables here since they describe the position of each point on the circle relative to the origin.
Fig. 1. A circle with radius r and centre (h, k), StudySmarter Originals
Using the distance formula between two points, we can calculate the distance between and as follows:
\[OP=\sqrt{(x-h)^2+(y-h)^2}\]
We can hereby introduce the term ‘radius’ as the distance between \((x,y)\) and the centre of the circle and denote it by \(r=OP\). Now, with the new symbol \(r\) for the radius of the circle, squaring both sides of the above equation, the square root is eliminated:
\[r^2=(x-h)^2+(y-k)^2\]
Which is none other than the equation we started with, using the definition of a circle. The equation obtained is the standard equation of a circle with centre and radius. The above form is particularly useful when the coordinates of the centre are given straightaway.
Give the equation of the circle whose radius is \((–1, –2)\) and radius is \(5\).
Solution
Recall the general form:
\[(x-h)^2+(y-k)^2=r^2\]
Where \((h, k)\) is the centre and \(r\) is the radius. Replacing \((h,k)\) with \((-1,-2)\) and \(r=5\), we get:
\[(x+1)^2+(y+2)^2=25\]
Hence the equation of the circle with radius \(5\) and centre \((–1, –2)\) is given by \((x+1)^2+(y+2)^2=25\).
Equation of a circle in the general form
Suppose we are given an equation where all the terms of the equation are expanded and \(h\), \(k\) cannot be deduced straightaway. In that case, we further build upon the obtained equation of a circle and derive another form of it, which is more general than the one above.
Expanding the previous equation, it is reduced to:
\[x^2-2xh-h^2+y^2-2yk+k^2=r^2\]
which can be rearranged as a standard quadratic with squared terms first, followed by the linear terms and then the constant:
\[x^2+y^2-2xh-2xk+h^2+k^2=r^2\]
To differentiate and avoid the conflict of constants between this equation and the former one, we introduce a set of new constants: \(h=-a\), \(k=-b\) and \(c=h^2+k^2-r^2\) to simplify the constant term.
After making these substitutions, we have the following equation of a circle in general form:
\[x^2+y^2+2ax+2by+c=0\]
The radius of the circle is now given by:
\[r^2=a^2+b^2-c\]
\[r=\sqrt{a^2+b^2-c}\]
Note that the condition \(a^2+b^2>c\) should be fulfilled, otherwise the radius will not be a positive real number and the circle won’t exist.
One can make little checks after solving an example, just to ensure that the answer makes sense, such as:
The coefficient of \(x^2\) and \(y^2\) should always be equal, if not then the equation does not describe a circle.
The inequality \(a^2+b^2>c\) is satisfied (otherwise, the radius is a complex number, which it cannot be).
It suffices for one of the conditions to not be met so that the answer at hand does not represent a circle.
One may also wonder how the equation of a circle can be constructed if we are given two points on it. The answer to that is that we cannot. There are an infinite number of circles passing through any two given points. In fact, to have a unique circle, at least three points on it should be known in order to find out its equation.
Equation of a Circle Centred at the Origin
The most common form of a circle will be a circle which is centred at the origin. In most cases, a circle is given and we can place our cartesian plane around it in such a way that it is easier to study its properties. And the most convenient place of setting our circle on a cartesian plane is centring it at the origin (since the centre is \((0,0)\) and calculations are much simpler).
Fig. 2.- A circle centred at the origin, StudySmarter Originals
Recall that the general form of a circle is given by:
\[(x-h)^2+(y-h)^2=r^2\]
Where \((h, k)\) represents the centre which can now be replaced with \((0,0)\):
\[x^2+y^2=r^2\]
Which is the Equation of a Circle centred at the origin.
Equation of a Circle given its Centre and a Point on the Circle
Suppose we are given not given the radius and centre of a circle, instead we are given a point on the circle \((x_1,y_1)\) and centre \((h,k)\). But the formula we have for the equation of the circle applies when the radius is known, hence we need to find the radius from the given data.
Going back to the definition of a circle, recall that radius is the distance between the centre and any point on the circle, here it is the distance between \((h,k)\) and \((x_1,y_1)\):
\[r^2=(x_1-h)^2+(y_1-k)^2\]
And since we know the general form as:
\[(x-h)^2+(y-k)^2=r^2\]
We can substitute for
\[r^2=(x_1-h)^2+(y_1-k)^2\]
Giving us:
\[(x-h)^2+(y-k)^2=(x_1-h)^2+(y_1-k)^2\]
Which is the equation of a circle whose centre is \((h,k)\) and \((x_1,y_1)\) lies on the circle.
Examples
Given that the radius of the circle \(x^2+y^2+2x+2y+k=0\) is \(5\), find the value of the real constant \(k\).
Solution:
Comparing the equation of the circle to the below general form:
\[x^2+y^2+2ax+2by+c=0\]
We can get the value of \(a\), \(b\) and \(c\):
\[2a=2,\quad 2b=2\]
\[a=1,\quad b=1\]
\[c=k\]
and the radius is given by \(r=\sqrt{a^2+b^2-c}\). And by substituting the values of \(a\), \(b\) and \(c\), we get
\[5=\sqrt{1^2+1^2-k}\]
\[k=-23\]
Hence the value of \(k\) is \(–23\).
Find the centre and radius of the circle \(x^2+y^2-2x-2y-2=0\) using both the methods: completing the square and the general form.
Solution:
Step 0: Verify if the given equation is a valid circle or not. We see that the coefficients of the squared terms are equal, thus it is a circle.
Method 1: Using the complete square method
Rearranging the \(x\) terms together and y terms together we get
\[x^2-2x+y^2-2y-2=0\]
Completing the square for \(x\) and \(y\), by adding and subtracting \(1\), we get
\[x^2-2x+1+y^2-2y+1-4=0\]
\[(x-1)^2+(y-1)^2=2^2\]
Comparing it to the \(h\), \(k\) form, it can be seen that the centre is \((1, 1)\) and the radius is \(2\).
Method 2: Using the general form
Comparing the given equation with the general form
\[x^2+y^2+2ax+2by+c=0\]
We get \(a=b=-1\) and \(c=-2\) where the centre has coordinates \((-a,-b)\) which converts to \((1,1)\) and the radius is
\[r=\sqrt{a^2+b^2-c}\]
\[r=\sqrt{1+1+2}=2\]
Thus the radius is \(2\) and centre is \((1,1)\).
As expected, the answer is the same using both methods.
A point relative to a circle
Suppose the coordinates of a random point are given to us and an equation of a circle is also given. We want to determine the position of the point with respect to the circle. And there are three possibilities:
the point is inside the circle;
outside the circle;
or on the circle.
There is no other scenario possible.
To determine where the point lies with respect to the circle, we need to look at the equation of the circle:
\[x^2+y^2+2ax+2by+c=0\]
If \(x^2+y^2+2ax+2by+c>0\), then the point \((x, y)\) lies outside the circle;
If \(x^2+y^2+2ax+2by+c<0\), then the point \((x, y)\) lies inside the circle;
If \(x^2+y^2+2ax+2by+c=0\), then the point \((x, y)\) lies on the circle (because it satisfies the equation of the circle).
To see why this is the case, recall the first standard form of the circle,
\[(x-h)^2+(y-k)^2=r^2\]
If the distance of the point from the centre is greater than the radius then it lies outside the circle. Similarly, if the distance is less than the radius of the circle then the point lies in the circle.
For the circle given by the equation \(x^2+y^2-4x+2y-1=0\), determine whether the points \(A(1,0)\) and \(B(2,-1)\) lie inside, outside or on the circle.
Solution:
For point \(A\), we evaluate the function at \((1, 0)\):
\[1+0-4+0-1=-4\]
\[-4<0\]
Hence, \(x^2+y^2-4x+2y-1<0\) at \(A\) which implies that point \(A\) lies inside the given circle.
For point \(B\), we follow the same procedure:
\[2^2+(-1)^2-4(2)-2-1=-6\]
\[-6<0\]
Thus, \(x^2+y^2-4x+2y-1<0\) for \(B\) and so the point \(B\) also lies inside the given circle.
Find the position of the point \((1,2)\) relative to the circle \(x^2+y^2+x-y+3=0\), i.e. determine whether it is inside, outside, or on the circle.
Solution:
We want to evaluate the function at \((1, 2)\),
\[1^2+2^2+1-2+3=7\]
\[7>0\]
Hence \(x^2+y^2+x-y+3>0\) at \((1,2)\) which implies that the point lies outside the circle.
Equation of a Circle - Key takeaways
- The equation of a circle when the centre \((h,k)\) and radius \(r\) are given is given by \((x-h)^2+(y-k)^2=r^2\).
- The general form (or the standard form) of a circle is given by \(x^2+y^2+2ax+2by+c=0\) where the centre of the circle is given by \((-a,-b)\) and the radius is given by \(r=\sqrt{a^2+b^2-c}\).
- For the circle \(x^2+y^2+2ax+2by+c=0\), a point lies outside the circle if \(x^2+y^2+2ax+2by+c>0\) at that point, inside the circle if \(x^2+y^2+2ax+2by+c<0\) and on the circle if \(x^2+y^2+2ax+2by+c=0\).
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