Explainer: Equation of a Circle Passing through Three Noncollinear Points

In this explainer, we will learn how to find the equation of a circle passing through three noncollinear points that form a right triangle.

You should know the two forms of the general equation of a circle.

Equation of a Circle in Radius-Center Form

The equation of a circle centered at 𝐢(β„Ž,π‘˜) and of radius π‘Ÿ is (π‘₯βˆ’β„Ž)2+(π‘¦βˆ’π‘˜)2=π‘Ÿ2.

Equation of a Circle in General Form

The equation of a circle can be written in the form π‘₯2+𝑦2+π‘Žπ‘₯+𝑏𝑦+𝑐=0.

How to Write the Equation of a Circle That Passes through Three Points in the Form (π‘₯ βˆ’ β„Ž)Β² + (𝑦 βˆ’ π‘˜)Β² = π‘ŸΒ²

  1. The coordinates of the three given points on the circle must satisfy the equation of the circle. With 𝑃1(π‘₯1,𝑦1), 𝑃2(π‘₯2,𝑦2), and 𝑃3(π‘₯3,𝑦3) the three given points, we can write (π‘₯1βˆ’β„Ž)2+(𝑦1βˆ’π‘˜)2=π‘Ÿ2,(π‘₯2βˆ’β„Ž)2+(𝑦2βˆ’π‘˜)2=π‘Ÿ2,(π‘₯3βˆ’β„Ž)2+(𝑦3βˆ’π‘˜)2=π‘Ÿ2.
    Note that given 𝑐=π‘Ÿ2, the above equations are simply saying that the distance between the center of the circle, of coordinates (β„Ž,π‘˜), and each of the three points is constant and equal to the radius.
  2. Since the three equations are in the form Expression =𝑐, they can be rearranged in a system of two equations with two unknowns by writing 𝐸1=𝐸2 and 𝐸1=𝐸3. (Note that any two combinations of the three equations work.) We get (π‘₯1βˆ’β„Ž)2+(𝑦1βˆ’π‘˜)2=(π‘₯2βˆ’β„Ž)2+(𝑦2βˆ’π‘˜)2,(π‘₯1βˆ’β„Ž)2+(𝑦1βˆ’π‘˜)2=(π‘₯3βˆ’β„Ž)2+(𝑦3βˆ’π‘˜)2.
    By expanding the brackets, we see that the terms β„Ž2 and π‘˜2 cancel out: π‘₯21βˆ’2π‘₯1β„Ž+𝑦21βˆ’2𝑦1π‘˜=π‘₯22βˆ’2π‘₯2β„Ž+𝑦2βˆ’2𝑦2π‘˜,π‘₯21βˆ’2π‘₯1β„Ž+𝑦21βˆ’2𝑦1π‘˜=π‘₯23βˆ’2π‘₯3β„Ž+𝑦3βˆ’2𝑦3π‘˜.
    By solving this system of equations, we find the coordinates of the circle center (β„Ž,π‘˜).
  3. The last stage is to plug in these values of β„Ž and π‘˜ in one of our first three equations to find the value of π‘Ÿ2.
  4. The equation of the circle is then (π‘₯βˆ’β„Ž)2+(π‘¦βˆ’π‘˜)2=π‘Ÿ2 with the values of β„Ž,π‘˜, and 𝑐 we have found.

Let’s see how this method is implemented when we have the coordinates of the three points.

Example 1: Writing the Equation of a Circle That Passes through Three Points in Center-Radius Form

Find the equation of the circle that passes through the points 𝐴(8,7), 𝐡(1,8), and 𝐢(0,1).

  1. (π‘₯βˆ’8)2+(π‘¦βˆ’8)2=10
  2. (π‘₯βˆ’4)2+(π‘¦βˆ’4)2=25
  3. (π‘₯+4)2+(𝑦+4)2=25
  4. π‘₯2+(π‘¦βˆ’1)2=50

Answer

  1. We are looking for the equation of the circle in the form (π‘₯βˆ’β„Ž)2+(π‘¦βˆ’π‘˜)2=π‘Ÿ2. Let’s write that the coordinates of the three points satisfy the equation of the circle: (8βˆ’β„Ž)2+(7βˆ’π‘˜)2=π‘Ÿ2,(1βˆ’β„Ž)2+(8βˆ’π‘˜)2=π‘Ÿ2,(0βˆ’β„Ž)2+(1βˆ’π‘˜)2=π‘Ÿ2.
  2. We arrange this system of three equations in a system of two equations: (8βˆ’β„Ž)2+(7βˆ’π‘˜)2=(1βˆ’β„Ž)2+(8βˆ’π‘˜)2,(8βˆ’β„Ž)2+(7βˆ’π‘˜)2=(0βˆ’β„Ž)2+(1βˆ’π‘˜)2.
    By expanding the brackets, we find 64βˆ’16β„Ž+β„Ž2+49βˆ’14π‘˜+π‘˜2=1βˆ’2β„Ž+64βˆ’16π‘˜+π‘˜2,64βˆ’16β„Ž+β„Ž2+49βˆ’14π‘˜+π‘˜2=β„Ž2+1βˆ’2π‘˜+π‘˜2.
    By rearranging and collecting like terms, we get βˆ’14β„Ž+2π‘˜=βˆ’48,βˆ’2β„Žβˆ’14π‘˜=βˆ’64.
    From the second equation, we get β„Ž=32+7π‘˜. By plugging in this value into the first equation, we find π‘˜=4. And by plugging this value of π‘˜ in one of the two equations above, we find β„Ž=4.
  3. By plugging in these values of β„Ž and π‘˜ in one of the first three equations, for instance, (0βˆ’β„Ž)2+(1βˆ’π‘˜)2=π‘Ÿ2, we have (0+4)2+(1βˆ’4)2=π‘Ÿ2, and thus π‘Ÿ2=25.
  4. The equation of the circle is: (π‘₯βˆ’4)2+(π‘¦βˆ’4)2=25.

Let’s see now another method to find the equation of a circle that passes through three points when we want to have the equation in the form π‘₯2+𝑦2+π‘Žπ‘₯+𝑏𝑦+𝑐=0.

How to Write the Equation of a Circle That Passes through Three Points in the Form π‘₯Β² + 𝑦² + π‘Žπ‘₯ + 𝑏𝑦 + 𝑐 = 0

  1. The coordinates of the three given points on the circle must satisfy the equation of the circle. With 𝑃1(π‘₯1,𝑦1), 𝑃2(π‘₯2,𝑦2), and 𝑃3(π‘₯3,𝑦3) the three given points, we can write π‘₯21+𝑦21+π‘Žπ‘₯1+𝑏𝑦1+𝑐=0,π‘₯22+𝑦22+π‘Žπ‘₯2+𝑏𝑦2+𝑐=0,π‘₯23+𝑦23+π‘Žπ‘₯3+𝑏𝑦3+𝑐=0.
  2. Solve the system of three equations to find the unknowns π‘Ž, 𝑏, and 𝑐.
  3. Plug the values found in the general equation π‘₯2+𝑦2+π‘Žπ‘₯+𝑏𝑦+𝑐=0.

Example 2: Writing the Equation of a Circle That Passes through Three Points in General Form

Determine the general equation of the shown circle 𝑀 passing through the origin point and the two points 𝐴(8,0) and 𝐡(0,βˆ’10).

  1. π‘₯2+𝑦2βˆ’16π‘₯+20𝑦=0
  2. π‘₯2+𝑦2βˆ’8π‘₯+10𝑦=0
  3. π‘₯2+𝑦2+10π‘₯βˆ’8𝑦=0
  4. π‘₯2+𝑦2+8π‘₯βˆ’10𝑦=0

Answer

  1. Here, we are asked to find the equation of the circle in the form π‘₯2+𝑦2+π‘Žπ‘₯+𝑏𝑦+𝑐=0. The coordinates of the three given points on the circle must satisfy the equation of the circle. We, therefore, plug the three pairs of coordinates in the general equation to get a system of three equations: 82+02+8π‘Ž+0𝑏+𝑐=0,02+(βˆ’10)2+0π‘Žβˆ’10𝑏+𝑐=0,02+02+0π‘Ž+0𝑏+𝑐=0.
  2. From the last equation, we find that 𝑐=0. This leaves us with 64+8π‘Ž=0,100βˆ’10𝑏=0.
    We find π‘Ž=βˆ’8,𝑏=10.
  3. The equation of the circle is π‘₯2+𝑦2βˆ’8π‘₯+10𝑦=0.

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