Equation of a circle

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:

1. The coefficient of \(x^2\) and \(y^2\) should always be equal, if not then the equation does not describe a circle.

2. 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

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:

1. the point is inside the circle;

2. outside the circle;

3. 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:

\[x^2+y^2+2ax+2by+c=0\]

1. If \(x^2+y^2+2ax+2by+c>0\), then the point \((x, y)\) lies outside the circle;

2. If \(x^2+y^2+2ax+2by+c<0\), then the point \((x, y)\) lies inside the circle;

3. 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.