In the previous post The Equations of The Tangent Lines to A Circle (1), it is discussed how to determine the equations of the tangent lines to a circle on condition that the slopes of the lines are given. Now, we will learn how to find such equations if the coordinates of the points of tangency are known.

**Case 2: The coordinates of the point of tangency is (x _{1},y_{1})**

In this case, the point with coordinates (x_{1},y_{1}) lies on the circle and it is the point of tangency. Suppose that the circle is centered at O(0,0) and its radius is R. Consequently, its equation is x^{2} + y^{2} = R^{2}. The equation of the tangent line at the point is as follows:

**x _{1}x + y_{1}y = R^{2} ** ……………………………………………………………………………………………………………………………………………………………………………………. (1)

**Example 1**

Let the equation of a circle be x^{2} + y^{2} = 5. A line touches the circle at (2,1). Find the equation of the corresponding tangent line.

**Answer**

The point with coordinates (2,1) lies on the circle. (See the important note at the end of this post.) Since it is the point of tangency, we can apply (1) to determine the equation of the corresponding tangent line. Substituting R^{2} = 5, x_{1} = 2 and y_{1} = 1 into (1), we get the line equation 2x + y = 5. So, the equation of the tangent line at (2,1) is **2x + y = 5**. (See Figure 1.)

**Figure 1**

**Example 2**

Find the equation of the tangent line to the circle x^{2} + y^{2} = 8 at the point whose abscissa is -2.

**Answer**

To find the ordinate of the point of tangency, substitute x = -2 into the equation of the circle. We thus get:

(-2)^{2} + y^{2} = 8

y^{2} – 4 = 0

(y – 2)(y + 2) = 0

There are two possible values of y, that is y = 2 or y = -2. So, in this example there are two points of tangency, i.e. P(-2,2) and Q(-2,-2).

To find the equation of the tangent line at P, substitute R^{2} = 8, x_{1} = -2, and y_{1} = 2 into (1). [We may apply (1) because it has been already known that P lies on the circle. See the important note at the end of this post.] It follows that:

-2x + 2y = 8

x – y = -4

To find the equation of the tangent line at Q, substitute R^{2} = 8, x_{1} = -2, and y_{1} = -2 into (1). [We may apply (1) because it has been already known that Q lies on the circle. See the important note at the end of this post.] This gives:

-2x – 2y = 8

x + y = -4

To sum up, there are two tangent lines at the points with abscissa -2, namely g_{1} and g_{2}, whose equations are determined as follows:

**g _{1} ≡ x – y = -4**

**g _{2} ≡ x + y = -4**

The tangent lines are depicted in Figure 2.

**Figure 2**

**Below is the equation of the tangent line to a circle centered at (α,β) if the coordinates of the point of tangency is (x _{1},y_{1}):**

**(x _{1} – α)(x – α) + (y_{1} – β)(y – β) = R^{2} **……………………………………………………………………………………………………………………………………….. (2)

**Example 3**

A circle L of radius 5 is centered at (2,1). Find the equation of the tangent line to the circle that passes through the point P(5,5).

**Answer**

The equation of the circle centered at (2,1) whose radius is 5 is (x – 2)^{2} + (y – 1)^{2} = 5^{2}. Thus, we have:

L ≡ (x – 2)^{2} + (y – 1)^{2} = 25 ……………………………………………………………………………………………………………………………………………………………… (*)

By substituting x = 5 and y = 5 into (*), we can conclude that P lies on L. In this case, P is the point of tangency. As a consequence, we are allowed to apply (2) to find the equation of the tangent line. [See the important note at the end of this post.] Substituting α = 2, β = 1, x_{1} = 5, y_{1} = 5, and R^{2} = 25 into (2), we get:

(5 – 2)(x – 2) + (5 – 1)(y – 1) = 25

3(x – 2) + 4(y – 1) = 25

3x + 4y = 35

So, the equation of the tangent line that passes through P(5,5) is **3x + 4y = 35**. (See Figure 3 below.)

**Figure 3**

**Important Note**

To apply (1) or (2), we must first check whether or not the point with coordinates (x_{1},y_{1}) is on the given circle. If the point does not lie on the circle, neither (1) nor (2) gives the desired equation!