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Given two points $(x_1,y_1)$ and $(x_2,y_2)$, recall that their horizontal distance from one another is $\Delta x=x_2-x_1$ and their vertical distance from one another is $\Delta y=y_2-y_1$. (Actually, the word "distance'' normally denotes "positive distance''. $\Delta x$ and $\Delta y$ are signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs $|\Delta x|$ and $|\Delta y|$, as shown in figure 1.2.1. The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides: $$ \hbox{distance} =\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}. $$ For example, the distance between points $A(2,1)$ and $B(3,3)$ is $\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}$.
Figure 1.2.1. Distance between two points, $\Delta x$ and $\Delta y$ positive.
As a special case of the distance formula, suppose we want to know the distance of a point $(x,y)$ to the origin. According to the distance formula, this is $\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$.
A point $(x,y)$ is at a distance $r$ from the origin if and only if $\sqrt{x^2+y^2}=r$, or, if we square both sides: $x^2+y^2=r^2$. This is the equation of the circle of radius $r$ centered at the origin. The special case $r=1$ is called the unit circle; its equation is $x^2+y^2=1$.
Similarly, if $C(h,k)$ is any fixed point, then a point $(x,y)$ is at a distance $r$ from the point $C$ if and only if $\sqrt{(x-h)^2+(y-k)^2}=r$, i.e., if and only if $$ (x-h)^2+(y-k)^2=r^2. $$ This is the equation of the circle of radius $r$ centered at the point $(h,k)$. For example, the circle of radius 5 centered at the point $(0,-6)$ has equation $(x-0)^2+(y- -6)^2=25$, or $x^2+(y+6)^2=25$. If we expand this we get $x^2+y^2+12y+36=25$ or $x^2+y^2+12y+11=0$, but the original form is usually more useful.
Example 1.2.1 Graph the circle $x^2-2x+y^2+4y-11=0$. With a little thought we convert this to $(x-1)^2+(y+2)^2-16=0$ or $(x-1)^2+(y+2)^2=16$. Now we see that this is the circle with radius 4 and center $(1,-2)$, which is easy to graph. $\square$
Exercises 1.2
Ex 1.2.1 Find the equation of the circle of radius 3 centered at:
(answer)
Ex 1.2.2 For each pair of points $A(x_1,y_1)$ and $B(x_2,y_2)$ find (i) $\Delta x$ and $\Delta y$ in going from $A$ to $B$, (ii) the slope of the line joining $A$ and $B$, (iii) the equation of the line joining $A$ and $B$ in the form $y=mx+b$, (iv) the distance from $A$ to $B$, and (v) an equation of the circle with center at $A$ that goes through $B$.
| a) $A(2,0)$, $B(4,3)$ | d) $A(-2,3)$, $B(4,3)$ |
| b) $A(1,-1)$, $B(0,2)$ | e) $A(-3,-2)$, $B(0,0)$ |
| c) $A(0,0)$, $B(-2,-2)$ | f) $A(0.01,-0.01)$, $B(-0.01,0.05)$ |
(answer)
Ex 1.2.3 Graph the circle $x^2+y^2+10y=0$.
Ex 1.2.4 Graph the circle $x^2-10x+y^2=24$.
Ex 1.2.5 Graph the circle $x^2-6x+y^2-8y=0$.
Ex 1.2.6 Find the standard equation of the circle passing through $(-2,1)$ and tangent to the line $3x-2y =6$ at the point $(4,3)$. Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.) (answer)
In this explainer, we will learn how to find the equation of a circle using its center and a given point or the radius and vice versa. Let us first recall the precise definition of a circle in mathematical terms. A circle is the locus of points equidistant from a given point, called the center of the circle. This fixed distance between any point of the circle and its center is the radius of the circle. In other words, a circle is the set of all points, and only those, that are at a given distance from its center.
We note that although a circle can easily be plotted on the 𝑥𝑦-plane, it cannot be described as a function of the form 𝑦=𝑓(𝑥) since any one element of the domain can (generally) be associated with two elements of the range. In other words, we can always find two points on the circle that have the same 𝑥-coordinate.
However, there does exist a relationship between the 𝑥- and 𝑦-coordinates of all points on the circle: this is the equation of a circle. To understand this equation, let us first of all consider the most basic form of a circle: a circle that is centered at the origin of the coordinate plane.
This circle is the locus of points equidistant from the origin. The distance from any point 𝑀(𝑥,𝑦) on the circle to the origin is thus equal to the radius of the circle, 𝑟. The relationship between the 𝑥- and 𝑦-coordinates of all the points on the circle can then be found by forming a right triangle as shown in the diagram below, where the hypotenuse is a radius of the circle.
Applying the Pythagorean theorem to this triangle, we find |𝑥|+|𝑦|=𝑟.
This expression applies to any point on the circle. The absolute values can be removed here since |𝑎|=𝑎, for any value of 𝑎. This leads to the following definition.
A circle with center (0,0) and radius 𝑟 is described by the equation 𝑥+𝑦=𝑟.
As one might expect, this equation can be extended to circles with any given center. Specifically, if we consider a circle of radius 𝑟 centered at point 𝐶(ℎ,𝑘), this is all the points that are a distance 𝑟 from 𝐶(ℎ,𝑘). If we consider a general point 𝑀(𝑥,𝑦) on the circle, we can form a right triangle between the center and this point in the same way as before, where the hypotenuse is the radius of the circle and the horizontal and vertical lengths are |𝑥−ℎ| and |𝑦−𝑘| respectively.
Using the Pythagorean theorem on this triangle, we find |𝑥−ℎ|+|𝑦−𝑘|=𝑟.
Once again, using the fact that |𝑎|=𝑎 for any 𝑎, we can rewrite this without the absolute values, leading to the following definition for the equation.
A circle with center (ℎ,𝑘) and radius 𝑟 is described by the equation (𝑥−ℎ)+(𝑦−𝑘)=𝑟.
An equation of this form is known as the standard form of the equation of a circle.
Let us consider how we can apply this equation to find the equation of a circle.
Write the equation of a circle with center (8,4) and radius 9.
Answer
Recall that the standard form for the equation of a circle is given by (𝑥−ℎ)+(𝑦−𝑘)=𝑟, where (ℎ,𝑘) is the center of the circle and 𝑟 is the radius.
In this example, we have been given that the center is (8,4), so ℎ=8 and 𝑘=4. We have also been given that the radius is 9, so 𝑟=9, and from this, we can get 𝑟=81. Substituting these values in the formula gives us the equation of the circle: (𝑥−8)+(𝑦−4)=81.
Just as we can find the equation of a circle if we are given its radius and center, so too can we determine the center and radius if we are just given the equation. Let us see how this is done below.
Find the center and radius of the circle (𝑥−2)+(𝑦−8)−100=0.
Answer
Let us remember that for a circle with center (ℎ,𝑘) and radius 𝑟, the standard form of its equation is (𝑥−ℎ)+(𝑦−𝑘)=𝑟.
We have been given the equation of a circle that is almost in this form, although the constant term is on the wrong side of the equation. Adding 100 to both sides gets it into this form, giving us (𝑥−2)+(𝑦−8)=100.
Comparing this to the general form of the equation, we can see that ℎ=2,𝑘=8,𝑟=100.
This means that the center is (2,8), and the radius 𝑟=√100=10.
Although the equation of a circle we have seen so far is the standard form that is used, a more general form of the equation exists. Let us recall that the standard form is (𝑥−ℎ)+(𝑦−𝑘)=𝑟.
If we expand the parentheses, this gives us 𝑥−2ℎ𝑥+ℎ+𝑦−2𝑘𝑦+𝑘=𝑟.
Rearranging this a bit so the terms with higher powers of 𝑥 and 𝑦 are on the left, we get 𝑥+𝑦−2ℎ𝑥−2𝑘𝑦+ℎ+𝑘−𝑟=0.
Notice that −2ℎ, −2𝑘 and ℎ+𝑘−𝑟 are all constants, so they can be written as 𝑎, 𝑏, and 𝑐 respectively. This results in the following equation.
The general form of the equation of a circle is 𝑥+𝑦+𝑎𝑥+𝑏𝑦+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants.
We note that this form does not directly involve the center and radius in the expression; instead, if we are given the center and the radius of a circle and want the general form, we must first write the equation in its standard form and then expand the parentheses. Let us see an example of this.
Give the general form of the equation of the circle of center (8,−2) and diameter 10.
Answer
Recall that the general form of the equation of a circle is 𝑥+𝑦+𝑎𝑥+𝑏𝑦+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants that need to be determined. To write the equation of a circle in this form, we can begin by writing it in its standard form and expanding the parentheses. The standard form is (𝑥−ℎ)+(𝑦−𝑘)=𝑟, where (ℎ,𝑘) is the center of the circle and 𝑟 is the radius.
In this instance, the circle has center (8,−2) and diameter 10. Since the diameter is twice the radius, the radius is 5. Thus, ℎ=8, 𝑘=−2, and 𝑟=5. Substituting these in, we get the following standard form of the equation: (𝑥−8)+(𝑦+2)=5=25.
We now want to get the equation into the general form, which we can achieve by expanding the parentheses. This gives us 𝑥−16𝑥+64+𝑦+4𝑦+4=25.
Finally, we can rearrange this to get the required form: 𝑥+𝑦−16𝑥+4𝑦+(64+4−25)=0𝑥+𝑦−16𝑥+4𝑦+43=0.
Just as we demonstrated that the standard form of the equation of a circle can give us the center and radius, we can imagine that it is also possible to find the center and radius of a circle given the general form. This is indeed the case; however, as this involves undoing the binomial expansion, we therefore need to be able to factor the equation by completing the square. Let us detail this procedure.
Suppose that we are given the equation of a circle in the general form 𝑥+𝑦+𝑎𝑥+𝑏𝑦+𝑐=0 and want to find the center and radius of the circle. We can do this by completing the square.
- First, we rearrange the equation as follows 𝑥+𝑎𝑥+𝑦+𝑏𝑦=−𝑐.
- Recall that we can complete the square by using a substitution of the form 𝑥+𝑎𝑥=𝑥+𝑎2−𝑎4. Completing the square for both parentheses in the equation gives us 𝑥+𝑎2−𝑎4+𝑦+𝑏2−𝑏4=−𝑐.
- Next, we can rearrange this so the constant terms are all on the right-hand side to get 𝑥+𝑎2+𝑦+𝑏2=𝑎4+𝑏4−𝑐.
- As this is the standard form of the equation of a circle, this tells us that the center is at −𝑎2,−𝑏2 and the radius is 𝑎4+𝑏4−𝑐.
Let us apply this procedure to find the center and radius for a given equation of a circle in general form.
By completing the square, find the center and radius of the circle 𝑥+6𝑥+𝑦−4𝑦+8=0.
Answer
As specified in the question, we are asked to find the center and radius of a circle given its equation in general form by completing the square. To do this, we can use the following substitutions: 𝑥+6𝑥=(𝑥+3)−9,𝑦−4𝑦=(𝑦−2)−4.
Putting these into the given equation, we get (𝑥+3)−9+(𝑦−2)−4+8=0.
Taking the constants to the other side, this is (𝑥+3)+(𝑦−2)=5.
This is the standard form for the equation of a circle. In other words, we have (𝑥−ℎ)+(𝑦−𝑘)=𝑟, where (ℎ,𝑘) is the center and 𝑟 is the radius of the circle. Therefore, the center is (−3,2), and the radius is 𝑟=√5.
Up until now, we have either been given the center and radius of a circle and have to write the equation or vice versa. Sometimes, however, we are not explicitly given all the information we need and instead have to work it out first through deduction. Let us consider an example of this.
A circle has center (2,2) and goes through the point (6,3). Find the equation of the circle.
Answer
Recall that the equation of a circle in its standard form is (𝑥−ℎ)+(𝑦−𝑘)=𝑟, where (ℎ,𝑘) is the circle’s center and 𝑟 is the radius.
In this example, we have been given the center but not the radius. If we substitute just ℎ=2 and 𝑘=2 into the above formula and leave 𝑟 as an unknown, we get (𝑥−2)+(𝑦−2)=𝑟.
Even though we do not have the radius, we know that any point on the circle has to satisfy this equation. Thus, if we put the point (6,3) into the above equation, we should get the correct value for 𝑟. Doing this gives us (6−2)+(3−2)=4+1=16+1=17.
Therefore, 𝑟=17, and the complete equation is (𝑥−2)+(𝑦−2)=17.
An alternative way to complete the above example is to find the radius by calculating the distance between the center and the given point. That is to say, if (6,3) is a point on the circle, this means that its distance from the center point (2,2) is equal to the radius. Recall that the distance between two points, (𝑥,𝑦) and (𝑥,𝑦), is given by the formula 𝑑=(𝑥−𝑥)+(𝑦−𝑦).
If we put the points (6,3) and (2,2) into this equation, we get (6−2)+(3−2)=√4+1=√16+1=√17.
This tells us the radius is √17; hence, 𝑟=17. This is equivalent to the above method because the standard equation of a circle is in essence just an equation for the distance between the center and a variable point. So, regardless of whether we calculate the distance directly or substitute a point into the equation, we are doing the same calculation.
For our final example, we will demonstrate what happens when we are given neither the radius nor the center of the circle but can deduce this information to help us find the equation.
Determine the general form of the equation of the circle that passes through the two points 𝐴(−7,1) and 𝐵(0,6), given that the circle’s center lies on the straight line 6𝑥−𝑦=−43.
Answer
Recall that the general form of the equation of the circle is given by 𝑥+𝑦+𝑎𝑥+𝑏𝑦+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants that need to be determined. To get to this form, we need to find the center of the circle and its radius, but we have not been given either.
Let us analyze the information we have been given and see how we can use it to solve the problem. We may not have the center of the circle, but we have two points that lie on the circle, and we know that all points on the circle are equidistant from the center. Let us suppose the center 𝐶 is (𝑥,𝑦). Since the distances from the center to each of the points (−7,1) and (0,6) are equal, this means we have the following equation: (𝑥+7)+(𝑦−1)=𝑥+(𝑦−6), where the left-hand side is the (squared) distance from (𝑥,𝑦) to (−7,1) and the right-hand side is the (squared) distance from (𝑥,𝑦) to (0,6). By expanding the parentheses, we get 𝑥+14𝑥+49+𝑦−2𝑦+1=𝑥+𝑦−12𝑦+36.
From here, we notice that the 𝑥 and 𝑦 terms cancel out. Thus, rearranging everything to the left-hand side, we get 14𝑥+10𝑦+14=0, and dividing by 2, we get 7𝑥+5𝑦+7=0.
Note that this equation tells us that point C lies on the line of equation 7𝑥+5𝑦+7=0. That is to be expected since the set of points that are equidistant to two distinct points forms a line that bisects the two points. We show this line below.
Now, we know that the center of the circle must lie on this line. By itself, this would not be enough to solve the problem, but recall that we have also been given that the center lies on a different line, 6𝑥−𝑦=−43. Assuming the lines are not parallel, they should intersect at exactly one point, which is where the center must be. We show this below.
We can find the intersection by substitution (i.e., we rearrange the second line (6𝑥−𝑦=−43) to be in terms of 𝑦): 6𝑥−𝑦=−43𝑦=6𝑥+43 and substitute this value for 𝑦 into the line 7𝑥+5𝑦+7=0 to get 7𝑥+5(6𝑥+43)+7=07𝑥+30𝑥+215+7=037𝑥=−222𝑥=−6.
Thus, the 𝑥-coordinate of the center is −6. We can find the 𝑦-coordinate by substitution too. Putting the value for 𝑥 into 𝑦=6𝑥+43, we get 𝑦=6(−6)+43=7.
Thus, the center is at (−6,7). Next, we need to find the radius. We can find this by taking the distance from (−6,7) to either point (let us just choose (0,6)). This is 𝑟=(−6)+(7−6)=36+1=37.
With this, we can now form the standard equation for a circle, which we can recall is (𝑥−ℎ)+(𝑦−𝑘)=𝑟, where (ℎ,𝑘) is the center and 𝑟 is the radius. Substituting in ℎ=−6, 𝑘=7, and 𝑟=37, this is (𝑥+6)+(𝑦−7)=37.
The final part of this question is to write the equation in its general form. We can do this by expanding the parentheses and rearranging, giving us 𝑥+12𝑥+36+𝑦−14𝑦+49=37𝑥+𝑦+12𝑥−14𝑦+48=0.
Although not required, plotting this equation gives us the following diagram.
Let us finish by considering the key points we have learned in this explainer.
- A circle with center (ℎ,𝑘) and radius 𝑟 has the following equation (in standard form): (𝑥−ℎ)+(𝑦−𝑘)=𝑟.
- The general form of the equation of a circle is 𝑥+𝑦+𝑎𝑥+𝑏𝑦+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants.
- The general form can be obtained by expanding the parentheses in the standard form.
- To find the center and radius from an equation in general form, we can complete the square to factor the equation into standard form.
- In problems where we are not given the center or radius, we can find these values by deduction and using the properties of a circle.