If a line and a circle are placed on a plane, there are three possibilities in terms of the number of points of intersection between them. First, the line intersects the circle at two distinct points. (See Figure 1 below.)

Figure1

Second, the line touches the circle, that is, the line intersects the circle at one and only one point. (Figure 2). The point where the line touches the circle is called the point of tangency. The line that touches the circle is called a tangent line.

Figure 2

Third, the line does not intersect the circle. (Figure 3)

Figure 3

Suppose that the equation of the circle is given, namely x2 + y2 = R2. [This circle is centered at the origin of the Cartesian plane and has radius R.] What are the equations of tangent lines to the circle?

 

Case 1: The slope of the tangent line is given

If the slope of the tangent line is m, then the equation of the tangent line is:

y = mx  \pm  R \sqrt{1 + m^2}      ……………………………………………………………………………………………………………………………………………………………….. (1)

 

Example 1

Given the circle having the equation x2 + y2 = 9, find the equation of the tangent line whose slope is 2. Also, find the coordinates of the points of tangency!

Answer

The equation of the circle can be rewritten as x2 + y2 = 32. From this, we know the radius of the circle, which is R = 3. To find the equation of the tangent line whose slope is 2, substitute m = 2 and R = 3 into (1). Consequently,

y = 2x \pm 3 \sqrt{1 + 2^2}

y = 2x \pm 3 \sqrt{5}

There are two lines that touch the circle, namely g1 and g2, which are described by the equations:

g1 : y = 2x + 3 \sqrt{5}
g2 : y = 2x - 3 \sqrt{5}

Figure 4

In Figure 4, P is the point where the line g1 touches the circle and Q is the point where the line g2 touches the circle. To determine the coordinates of P, substitute y in the equation of g1, i.e. y = 2x + 3√5, into the equation of the circle x2 + y2 = 9. It follows that:

x^2 + (2x + 3 \sqrt{5})^2 = 9

5x^2 + 12x \sqrt{5} + 36 = 0

This quadratic equation has a double root x = - \frac{6}{5} \sqrt{5}. To obtain the ordinate of P, replace x in the equation of g1 with - \frac{6}{5} \sqrt{5}. As a consequence:

y = 2(- \frac{6}{5} \sqrt{5}) + 3 \sqrt{5}

y = \frac{3}{5} \sqrt{5}

Thus, P(- \frac{6}{5} \sqrt{5}, \frac{3}{5} \sqrt{5})

To determine the coordinates of Q, substitute y in the equation of g2, i.e. y = 2x – 3√5 into the equation of the circle x2 + y2 = 9. It follows that:

x^2 + (2x - 3 \sqrt{5})^2 = 9

5x^2 - 12x \sqrt{5} + 36 = 0

This quadratic equation has a double root x = \frac{6}{5} \sqrt{5}. To obtain the ordinate of Q, replace x in the equation of g2 with \frac{6}{5} \sqrt{5}. As a consequence:

y = 2(\frac{6}{5} \sqrt{5}) - 3 \sqrt{5}

y = - \frac{3}{5} \sqrt{5}

Therefore, Q(\frac{6}{5} \sqrt{5}, - \frac{3}{5} \sqrt{5})

 

Below is the equation of the tangent line to a circle centered at (α,β) if the slope of the tangent line is m:

y - \beta = m(x - \alpha) \pm  R \sqrt{1 + m^2} ……………………………………………………………………………………………………………………………………………. (2)

 

Example 2

Let L be a circle centered at (-4,1) with radius 3. The lines g1 and g2 touch L and have a slope of 2. Find the equations of the tangent lines.

Answer

In this case, α = -4, β = 1, R = 3, and m = 2. Substituting the values into (2), we get:

y - 1 = 2(x - (-4)) \pm 3 \sqrt{1 + 2^2}

y = 2x + (9 \pm 3 \sqrt{5})

Thus, the equations of g1 and g2 are:

g1 : y = 2x + (9 + 3 \sqrt{5})
g2 : y = 2x + (9 - 3 \sqrt{5})

Figure 5

 

Case 2: The coordinates of the point of tangency is given

This case is elaborated in the article The Equations of The Tangent Lines to A Circle (2).

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