THE EQUATION OF THE POLAR OF A POINT WITH RESPECT TO A CIRCLE (2)

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(x₁,y₁) with respect to the circle (x – α)² + (y – β)² = R² is:

(x₁ – α)(x – α) + (y₁ – β)(y – β) = R²

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 = 2²

(x + 1)² + (y – 3)² = 4

In this example, we have: α = -1, β = 3, x₁ = -4, and y₁ = 4.

Substitute these values into the equation (x₁ – α)(x – α) + (y₁ – β)(y – β) = R² 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 x² + y² + Ax + By + C = 0 and the pole has coordinates (x₁,y₁) what is the equation of the corresponding polar?

Example 2

Given the circle x² + y² + 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, x₁ = -4, and y₁ = 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 (x₁,y₁) 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 (x₁,y₁).

 

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

 

Example 3

Given the circle x² + y² – 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 x₁ = 3, y₁ = 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 x² + y² = 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 x² + y² – 36 = 0.

In this example, A = B = 0, C = – 36, x₁ = 8, y₁ = 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 (x₁,y₁) and (x₂,y₂) is . Applying the formula, we can obtain the equations of the tangent lines, i.e. g₁ and g₂ in Figure 3, as follows.