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.