In the article The Equation of The Polar of A Point with Respect to A Circle (1), it is shown that the equation of the polar of the point K(x_{1},y_{1}) with respect to the circle (x – α)^{2} + (y – β)^{2} = R^{2} is:

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

This post is intended to give some examples of how it is applied and indicate the interplay between the polar and the tangent lines.

**Example 1**

A circle has center at (-1.3) and radius 2. Find the equation of the polar of the point/pole whose coordinates is (-4.4).

**Answer**

The equation of the circle with center at (-1.3) and radius 2 is:

(x + 1)^{2} + (y – 3)^{2} = 2^{2}

(x + 1)^{2} + (y – 3)^{2} = 4

In this example, we have: α = -1, β = 3, x_{1} = -4, and y_{1} = 4.

Substitute these values into the equation (x_{1} – α)(x – α) + (y_{1} – β)(y – β) = R^{2} above, resulting in:

(-4 – (-1))(x – (-1)) + (4 – 3)(y – 3) = 22

-3(x + 1) + (y – 3) = 4

-3x + y = 10

**y = 3x + 10** [This is the desired equation.]

(See Figure 1)

**Figure 1**

Now, if the equation of the circle is of the form x^{2} + y^{2} + Ax + By + C = 0 and the pole has coordinates (x_{1},y_{1}) what is the equation of the corresponding polar?

**Example 2**

Given the circle x^{2} + y^{2} + 2x – 6y + 6 = 0, determine the equation of the polar of the pole whose coordinates is (-4,4).

**Answer**

In this example, A = 2, B = -6, C = 6, x_{1} = -4, and y_{1} = 4.

Substituting these values into (*) , we have the following:

Simplifying the equation, we obtain **y = 3x + 10**. [Compare with the result in Example 1.]

**Note:**

The circles in Example 1 and Example 2 are identical, but have different forms of equations. That’s why they give the same polar.

**What if we apply (*) while (x _{1},y_{1}) is the coordinates of a point that lies on the circle? In this case, (*) will result in the equation of the tangent line that touches the circle at (x_{1},y_{1}).**

Example 3 below shows the application of (*) if the point under consideration lies on the circle.

**Example 3**

Given the circle x^{2} + y^{2} – 4x – 6y + 8 = 0, find the equation of the tangent line that passes through the point (3,5).

**Answer**

By substituting x = 3 and y = 5 into the equation of the circle, it can be concluded that the point (3.5) lies on the circle, so we can use (*) to determine the equation of the tangent line. Substituting x_{1} = 3, y_{1} = 5, A = -4, B = -6, and C = 8 into (*), we have the following:

By simplifying the equation, the desired equation can be expressed as **x + 2y = 13**. (See Figure 2.)

**Figure 2**

The following example shows the relationship between the polar and the corresponding tangent lines.

**Example 4**

Given the circle x^{2} + y^{2} = 36, find the equations of the tangent lines passing through the point K(8,0).

**Answer**

[Since this post is about polars, the solution to this problem will be sought through the polar of K with respect to the circle. Actually this can be solved without polars.]

The equation of the circle can be expressed as x^{2} + y^{2} – 36 = 0.

In this example, A = B = 0, C = – 36, x_{1} = 8, y_{1} = 0. Substituting these values into (*), we get the equation of the polar 8x – 36 = 0, which is equivalent to x = 4½. In Figure 3, the polar is drawn in red. To get the points of tangency, substitute x = 4½ into the equation of the circle. This gives . So, the coordinates of the points of tangency are (4½,1½√7) and (4½,-1½√7).

**Figure 3**

The equation of the line passing through the points (x_{1},y_{1}) and (x_{2},y_{2}) is . Applying the formula, we can obtain the equations of the tangent lines, i.e. g_{1} and g_{2} in Figure 3, as follows.