Lesson Explainer: Equation of a Circle Mathematics

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.

How To: Describing a Circle Mathematically

Mathematically, a circle can be described as the locus of points equidistant from a given point, called the circle’s center. It means that the circle is the set of all the points, and only those, that are at a given distance from the center of the circle. This fixed distance between any point of the circle and its center is the radius of the circle.

Note that a circle is not the graph of a function 𝑦=𝑓(π‘₯) since one element of the domain can be associated with two elements of its range. In other words, we can find two points on the circle that have the same π‘₯-coordinate.

However, there exists a relationship between the π‘₯- and 𝑦-coordinates of all the points on the circle: this is the equation of a circle.

Equation of a Circle Centered at the Origin in Center-Radius Form

Let’s start with a circle 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(π‘₯,𝑦) on the circle to the origin is the circle radius, π‘Ÿ. The relationship between the π‘₯- and 𝑦-coordinates of all the points on the circle is then given by applying the Pythagorean theorem in the right triangle shown in the diagram below, where the hypotenuse is a radius of the circle.

We find |π‘₯|+|𝑦|=π‘Ÿ.

The absolute values can be removed since they are squared (|π‘₯|=π‘₯, whatever the sign of π‘₯). Therefore, π‘₯+𝑦=π‘Ÿ.

This is the equation of a circle of radius π‘Ÿ centered at the origin.

We will now find the equation of any circle.

Equation of a Circle of Radius π‘Ÿ Centered at 𝐢(β„Ž,π‘˜) in Center-Radius Form

The circle of radius π‘Ÿ centered at 𝐢(β„Ž,π‘˜) represents the locus of points equidistant from point 𝐢(β„Ž,π‘˜). Any point on the circle is at a distance π‘Ÿ from the center 𝐢(β„Ž,π‘˜).

We apply the Pythagorean theorem in the right triangle shown in the diagram below, where the hypotenuse is a radius of the circle.

We find |π‘₯βˆ’β„Ž|+|π‘¦βˆ’π‘˜|=π‘ŸοŒ¬οŠ¨οŒ¬οŠ¨οŠ¨ which can be rewritten as (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘Ÿ.

This is true for any point on the circle, so the equation of a circle of radius π‘Ÿ centered at 𝐢(β„Ž,π‘˜), which describes the relationship between the π‘₯- and 𝑦-coordinates of all the points on the circle, can be written as (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘Ÿ.

Note that the general equation of a circle can also be derived from the equation of a circle of radius π‘Ÿ centered at the origin by translating the circle β„Ž units horizontally and π‘˜ units vertically, that is, by vector (β„Ž,π‘˜).

The equation of the circle given above is written in the so-called center-radius form. The equation of a circle can be written in another form, called the general form. This form is obtained simply by expanding the brackets in the equation in center-radius form.

Equation of a Circle in General Form

The equation of a circle of radius π‘Ÿ centered at 𝐢(β„Ž,π‘˜) is (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨. By expanding the brackets, we get π‘₯+π‘¦βˆ’2β„Žπ‘₯βˆ’2π‘˜π‘¦+β„Ž+π‘˜=π‘Ÿ. This can be rewritten as π‘₯+π‘¦βˆ’2β„Žπ‘₯βˆ’2π‘˜π‘¦+β„Ž+π‘˜βˆ’π‘Ÿ=0.

Let βˆ’2β„Ž be π‘Ž, βˆ’2π‘˜ be 𝑏, and β„Ž+π‘˜βˆ’π‘ŸοŠ¨οŠ¨οŠ¨ be 𝑐; we get π‘₯+𝑦+π‘Žπ‘₯+𝑏𝑦+𝑐=0.

This is the equation of a circle in general form.

Example 1: Writing the Equation of a Circle Given Its Center

Write, in the form π‘Žπ‘₯+𝑏𝑦+𝑐π‘₯+𝑑𝑦+𝑒=0, the equation of the circle of radius 10 and center (4,βˆ’7).

Answer

We start by writing the equation of a circle: (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘Ÿ.

The radius π‘Ÿ is 10 and the center coordinates are β„Ž=4 and π‘˜=βˆ’7, so this gives us (π‘₯βˆ’4)+(𝑦+7)=10(π‘₯βˆ’4)+(𝑦+7)=100.

This is the equation of the circle of radius 10 and center (4,βˆ’7) in center-radius form.

However, we are asked to give it in the form π‘₯+𝑦+π‘Žπ‘₯+𝑏𝑦+𝑐=0.

We need to expand the brackets, π‘₯βˆ’8π‘₯+16+𝑦+14𝑦+49=100, and then take away 100 from each side, π‘₯βˆ’8π‘₯+16+𝑦+14𝑦+49βˆ’100=0, and collect like terms: π‘₯+π‘¦βˆ’8π‘₯+14π‘¦βˆ’35=0.

Example 2: Writing the Equation of a Circle Given Its Center

In the figure below, find the equation of the circle.

Answer

In this example, we need to use the graph to identify the center’s coordinates and the radius of the circle.

The circle center’s coordinates are (β„Ž,π‘˜)=(βˆ’5,βˆ’4).

To find the radius, we can, for instance, work out the difference in the 𝑦-coordinates of the highest point and the center, 1βˆ’(βˆ’4)=1+4=5, or the difference in the π‘₯-coordinates of the furthest point to the right and the center: 0βˆ’(βˆ’5)=5. So π‘Ÿ=5.

We plug the values of β„Ž, π‘˜, and π‘Ÿ in (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨ and find (π‘₯+5)+(𝑦+4)=25.

Example 3: Writing the Equation of a Circle Given Its Center

Determine the equation of a circle that passes through the point 𝐴(0,8) if its center is 𝑀(βˆ’2,βˆ’6).

Answer

We start by writing the general equation of a circle: (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘Ÿ.

We know that point 𝑀(βˆ’2,βˆ’6) is the center of the circle, so β„Ž=βˆ’2 and π‘˜=βˆ’6. We then plug in these values in the equation, and we get (π‘₯+2)+(𝑦+6)=π‘Ÿ.

We do not know the radius, but we know that point 𝐴 is on the circle, so its coordinates π‘₯=0 and 𝑦=8 must satisfy the equation of the circle. We can, therefore, substitute π‘₯ and 𝑦 in the equation with these values to find π‘Ÿ: (2)+(8+6)=π‘Ÿ4+196=π‘Ÿ200=π‘Ÿ.

The equation of the circle is finally (π‘₯+2)+(𝑦+6)=200.

How To: Finding the Center Coordinates and the Radius from the Equation in the Center–Radius Form

Given the equation of a circle in the form (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨, the center coordinates are (β„Ž,π‘˜) and the radius is π‘Ÿ=βˆšπ‘ŸοŠ¨.

Example 4: Finding the Center Coordinates and the Radius of a Circle from Its Equation in Center-Radius Form

Find the center and radius of the circle (π‘₯βˆ’2)+(𝑦+8)βˆ’100=0.

Answer

  1. We need to rearrange the equation in the form (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨. We get (π‘₯βˆ’2)+(𝑦+8)=100.
  2. By comparing the given equation with (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨, we find that β„Ž=2, π‘˜=βˆ’8, and π‘Ÿ=100.
  3. The center coordinates are (2,βˆ’8) and the radius is π‘Ÿ=βˆšπ‘Ÿ=√100=10.

How To: Finding the Center Coordinates and the Radius from the Equation in the General Form

When the equation of a circle is given in the general form π‘₯+𝑦+𝑏π‘₯+𝑐𝑦+𝑑=0, the equation should be rewritten in the form (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨ by completing the square for π‘₯+𝑏π‘₯ and 𝑦+π‘π‘¦οŠ¨.

This gives ο€½π‘₯+𝑏2+𝑦+𝑐2=π‘ŸοŠ¨οŠ¨οŠ¨, which allows the identification of the center (β„Ž,π‘˜)=ο€½βˆ’π‘2,βˆ’π‘2 and radius π‘Ÿ=βˆšπ‘ŸοŠ¨ of the circle.

Example 5: Finding the Center Coordinates and the Radius of a Circle from Its Equation in Standard Form

By completing the square, find the center and radius of the circle π‘₯+6π‘₯+π‘¦βˆ’4𝑦+8=0.

Answer

  1. We need to rearrange the equation in the form (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨ by completing the square.
  2. We find π‘₯+6π‘₯=(π‘₯+3)βˆ’9 and π‘¦βˆ’4𝑦=(π‘¦βˆ’2)βˆ’4.
  3. By substituting these in the original equation, we get (π‘₯+3)βˆ’9+(π‘¦βˆ’2)βˆ’4+8=0.
  4. By rearranging in the form (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨, we find (π‘₯+3)+(π‘¦βˆ’2)=5.
  5. We find that β„Ž=βˆ’3, π‘˜=2, and π‘Ÿ=5.
  6. The center coordinates are (βˆ’3,2) and the radius is π‘Ÿ=βˆšπ‘Ÿ=√5.

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