This lesson will cover a few examples to illustrate the concept covered in the previous lesson: position of a point with respect to a circle.
Example 1. Determine the position of the following points with respect to the given circles
(i) (3, 4) with respect to x2 + y2 – 9 = 0
(ii) (1, 0) with respect to x2 + y2 – 2x + 4y + 1 = 0
(iii) (2, 3) with respect to x2 + y2 – 6x – 4y = 0
Solution. Recall that all we need to do is evaluate S1, i.e. substitute the coordinates of the given points in the corresponding circles’ equations, and check its sign. Positive means outside, negative means inside, and zero means on the circle.
(i) Here S1 = 32 + 42 – 9 = 16, which is positive. Therefore the point lies outside the given circle.
(ii) In this case S1 = 12 + 02 – 2(1) + 4(0) + 1 = 0. Therefore the point lies on the given circle.
(iii) Here S1 = 22 + 32 – 6(2) – 4(3) = –11, which is negative. Therefore the point lies inside the given circle.
Example 2. Find the range of values of ‘a’, such that the point (a, 2a – 3) lies in the region common to the two circles x2 + y2 = 5, and x2 + y2 – 6x = 0, as shown.
Solution. For the point to lie in the common region for both the circles, the point must satisfy the conditions to lie inside both the circles. Do you agree?
That is, S1 for the point (a, 2a – 3) should be negative for both the circles. Let’s do that.
For the first circle, we have a2 + (2a – 3)2 – 5 < 0, or 5a2 – 12a + 4 < 0 or 2/5 < a < 2
For the second one, a2 + (2a – 3)2 – 6a < 0, or 5a2 – 18a + 9 < 0 or 3/5 < a < 3
By combining the two cases we get the required set of values for ‘a’ as 3/5 < a < 2
Example 3. Find the set of values of ‘a’ for which the point (a, a + 1) lies in the minor segment of the circle x2 + y2 = 25 formed by the line 2x + y – 4 = 0
Solution. Have a look at the following figure, to understand the problem better.
From the figure, it is clear that the point must lie inside the circle. But what will keep it in the minor segment?
Something to do with its position relative to the line. Rings a bell?
The point will lie in the minor segment if it’s on the opposite side of the given line, as that of the circle’s center.
Makes sense? Now let’s do the math.
For the point to lie inside the circle, S1 < 0. That is a2 + (a + 1)2 < 25, or – 4 < a < 3
And, for the point to lie on the opposite side of the center, we have (2(0) + (0) – 4)(2a + a + 1 – 4) < 0, or a > 1
Combining the two conditions, we finally have 1 < a < 3.
And that would be all for this lesson.
The next few lessons will cover intersection of a line with a circle.