Math Lesson 15.6.5 - What Else We Can Find Using the Equation of a Circle?

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Welcome to our Math lesson on What Else We Can Find Using the Equation of a Circle?, this is the fifth lesson of our suite of math lessons covering the topic of Circle Graphs, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

What Else We Can Find Using the Equation of a Circle?

We can find many more things than just the centre and the radius from the equation of a circle. For example, we can find the intercepts of the circle with any of the axes. There may be 0, 1 or 2 intercepts with each axis depending on the position and size of the circle. However, you don't need to draw a circle in the coordinates system to know the number of intercepts. You simply can substitute one of the coordinates by 0, as we have done in other cases and solve the remaining equation in terms of the other variable. This gives the intercept with the axis that contains this variable.

For example, if you want to find the X-intercept of the circle

you substitute y = 0 and solve the equation in terms of the variable x. In this way, you obtain

This equation is true for (x - 4) = √7 and -(x - 4) = √7. The first option gives

and the second option gives

Therefore, the x-intercepts are A(1.35, 0) and B(6.65, 0).

As for the Y-intercepts, we take x = 0. In this way, we obtain

Thus, there is a single Y-intercept of the circle. In this case, we say the circle is tangent with the Y-axis at D(0, 3), and intersecting with the X-axis at A(1.35, 0) and B(6.65, 0).

In other words, if the circle touches any of the axes at a single point, we say the circle is tangent with that axis, while if the circle intercepts an axis at two points, we say the circle is intersecting with that axis at these two points. Look at the figure below that shows the circle discussed above.

The third case includes situations where the circle doesn't have any contact with one or both axes. In this case, the one-variable equation produced when substituting one of the variables with 0 has no solutions. Let's illustrate this point with a couple of examples.

Example 4

Find the intercepts (if any) of the following circles with the X- and Y-axes.

1. (x + 2)² + (y - 1)² = 9
2. (y + 4)² + (y + 5)² = 9

Solution 4

1. (x + 2)² + (y - 1)² = 9
   Let's begin with the y-intercept. Thus, for x = 0, we have
   (0 + 2)² + (y - 1)² = 9
   4 + (y - 1)² = 9
   (y - 1)² = 5
   There are two y-intercepts:
   y - 1 = √5 and -(y - 1) = √5
   We have y = √5 + 1 ≈ 3.24 and y = 1 - √5 ≈ -1.24. Therefore, the two y-intercepts are (0, -1.24) and (0, 3.24).
   As for the x-intercepts, we have y = 0. Thus,
   (x + 2)² + (0 - 1)² = 9
   (x + 2)² + 1 = 9
   (x + 2)² = 8
   Thus, we have x + 2 = √8 and -(x + 2) = √8. The first equation gives x ≈ 0.83, while the second equation gives x = -4.83. Therefore, the x-intercepts are (-4.83, 0) and (0.83, 0).
   The graph below shows these four intercept points.
2. (x + 4)² + (y + 5)² = 9
   Let's begin with the y-intercept. Thus, for x = 0, we have
   (0 + 4)² + (y + 5)² = 9
   16 + (y + 5)² = 9
   (y + 5)² = 9 - 16
   (y + 5)² = -7
   This equation is always false for all values of y because no number raised into the second power can give a negative result. Therefore, this circle has no y-intercepts. As for the x-intercepts, we must take y = 0. Thus,
   (x + 4)² + (0 + 5)² = 9
   (x + 4)² + 25 = 9
   (x + 4)² = 9-25
   (x + 4)² = -16
   This equation is always false for all values of x because no number raised into the second power can give a negative result. Therefore, this circle has no x-intercepts.

Another thing you can find using the equation of a circle is the equation of the horizontal or vertical diameter. Thus, if you know the centre C(a, b) and the radius r of a circle (both are easily identified if the equation of the circle is given), then the equation of the horizontal diameter is y = b, which corresponds to the vertical coordinate of the centre C. Likewise, the vertical diameter has the equation x = a, given that it must pass through the horizontal coordinate of the centre C.

As for the allowed values, it is obvious that the diameter must extend horizontally within the range a - r ≤ x ≤ a + r and vertically it must extend within the range b - r ≤ x ≤ b + r, as shown in the figure below.

Example 5

1. Find the equation of the horizontal and vertical diameter for the circle shown in the figure.
2. Find the equation of this circle.

Solution 5

1. From the figure, it is evident that the centre C is at C(5, -3). Thus, we have a = 5 and b = -3. Therefore, the equation of the vertical diameter (without the restrictions) is x = a = 5 and that of the horizontal diameter (again, without the restrictions) is y = b = -3.
   Another thing we can see from the figure is the extension of the circle. In the horizontal direction, the circle extends between x_min = -1 and x_max = 11. Likewise, the vertical extension of this circle is between y_min = -9 and y_max = 3. Hence, the equations of the two diagonals are
   Horizontal diamater: y = -3 - 1 ≤ x ≤ 11
   and
   Vertical diamater: x = 5 - 9 ≤ y ≤ 3
2. Using the above findings, we obtain for the radius r:
   r = x_max - x_min/2
   = 11 - (-1)/2
   = 12/2
   = 6
   Therefore, since the general equation of a circle is
   (x - a)² + (y - b)² = r²
   in the specific case, we obtain
   (x - 5)² + (y - (-3))² = 6²
   (x - 5)² + (y + 3)² = 36

You have reached the end of Math lesson 15.6.5 What Else We Can Find Using the Equation of a Circle?. There are 5 lessons in this physics tutorial covering Circle Graphs, you can access all the lessons from this tutorial below.

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