2.8 F) Circles – Part 2
It may be the case in the exam that you are asked to find an equation of the tangent to a circle at a particular point. A tangent is a straight line that touches the circle. The tangent of a circle is perpendicular to the radius of the circle (perpendicular means that the tangent and radius meet at a right-angle/ 90°). This means that the gradient of the tangent to the circle at a particular point will be the negative reciprocal of the radius of the circle at that particular point. To find the negative reciprocal, we flip the number (the original numerator becomes the new denominator and the original denominator becomes the new numerator) and add a minus sign in front of it.
After finding the gradient of the tangent, we are able to find the equation of the tangent because we have the gradient of the tangent and a point that the line passes through. We find the equation of the tangent by replacing m in y = mx + c (the general linear form) and subbing in the x and y value for the particular point that the tangent passes through.
Let’s have a few examples.
A circle has a centre at the origin and the point (4, 3) lies on the circle. What is the equation of the tangent to the circle at the point (4, 3)?
Like the examples that we looked at in the previous section, the first step in answering this question is to create a quick sketch with all of the key points labelled so that we can visualise what is happening.
The next step is to work out what the gradient of the radius is, which we are able to do because we have 2 points that lie on the radius; (0, 0) and (4, 3). We work out the gradient of a line by using the formula below:
It does not matter what point we take as point 1 or point 2; we will get the same gradient either way. I am going to have point 1 as (0, 0) and point 2 as (4, 3). We now label the points up and sub the values into the gradient formula.
The gradient of the radius is 3/4. We know that the gradient of the tangent will be the negative reciprocal of the gradient of the radius. To find the negative reciprocal, we flip the fraction (the original numerator becomes the new denominator and the original denominator becomes the new numerator) and we add a minus. Therefore, the gradient of the tangent is -4/3.
The gradient that we have found is consistent with our sketch; the tangent on our sketch is downwards sloping, which means that it has a negative gradient and we have found a negative gradient.
We now have the gradient of the tangent and a point that lies on the tangent, which means that we are able to find an equation for the line. I am first going to replace m in the general equation with the gradient that we have found; I replace m with -4/3.
The next step in to sub in the x and y values from the coordinates into the equation. We are told that the tangent passes through (4, 3); we therefore sub x in as 4 and y in as 3.
In order to complete the calculation on the left, we need to make the denominators the same. At the moment the 3 can be written as 3/1. We want the denominator to be 3, so we multiply the numerator and the denominator of the fraction by 3. This results in 3 becoming 9/3. As the denominators are now the same, we can add the two numerators together and keep the denominator the same.
The final step is to make c a mixed number (currently the fraction is improper because the numerator is larger than the denominator). We convert the improper fraction into a mixed number by seeing how many times the denominator fully goes into the numerator; this gives us the number of wholes in the improper fraction. The remainder is then the numerator of the fraction part of the mixed number.
The next step is to turn the gradient into a mixed number.
Therefore, the equation of the tangent to the circle at the point (4, 3) is:
A circle has a centre at (0, 0) and the point (-2, 5) lies on the circle. What is the equation of the tangent to the circle at the point (-2, 5)?
Let’s draw a quick sketch.
Let’s work out the gradient of the radius of the circle, which we do by using the two points that we are told lie on the radius; (0,0) and (-2, 5).
The gradient of the radius is -5/2. The gradient of the tangent is the negative reciprocal to the gradient of the radius.
The gradient of the tangent is 2/5, which looks correct from our diagram because the tangent on our quick sketch is upwards sloping (positive gradient).
We are now able to find the equation of the tangent because we have the gradient of the line and a point that the line passes through. The gradient of the line is 2/5, which means that we replace m in the general linear equation form with 2/5.
The next step is to sub in the x and y values from the point where the tangent passes through. The tangent passes through the point (-2, 5), which means that we sub x in as -2 and y in as 5.
In order to complete the calculation on the left, we need to make the 5 into a fraction with the same denominator as the second fraction. Currently the 5 can be written as 5/1 and we want the denominator to become 5. We can make the denominator 5 by multiplying the numerator and the denominator by 5. This results in the 5 becoming 25/5. We are then able to complete the calculation by adding the numerators and keeping the denominator the same.
We would then change c into a mixed number.
Therefore, the equation of the tangent is:
We also could have written the equation of the line in decimal form: