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.

**Q1. **Determine the position of the following points with respect to the given circles

(i) (3, 4) with respect to x^{2} + y^{2} – 9 = 0

(ii) (1, 0) with respect to x^{2} + y^{2} – 2x + 4y + 1 = 0

(iii) (2, 3) with respect to x^{2} + y^{2} – 6x – 4y = 0

**Sol. **Recall that all we need to do is evaluate S_{1}, i.e. substitute the coordinates of the given points in the corresponding circles’ equations, and check its sign. Positive means outside, negative – inside, and zero – on the circle.

(i) Here S_{1} = 3^{2} + 4^{2} – 9 = 16, which is positive. Therefore the point lies **outside **the given circle.

(ii) In this case S_{1} = 1^{2} + 0^{2} – 2(1) + 4(0) + 1 = 0. Therefore the point lies **on** the given circle.

(iii) Here S_{1} = 2^{2} + 3^{2} – 6(2) – 4(3) = –11, which is negative. Therefore the point lies **inside** the given circle.

One more.

**Q2. **Find the range of values of ‘a’, such that the point (a, 2a – 3) lies in the region common to the two circles x^{2} + y^{2} = 5, and x^{2} + y^{2} – 6x = 0, as shown.

**Sol. **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, S_{1} for the point (a, 2a – 3) should be negative for both the circles. Let’s do that.

For the first circle, we have a^{2} + (2a – 3)^{2} – 5 < 0, or 5a^{2} – 12a + 4 < 0 or **2/5 < a < 2**

For the second one, a^{2} + (2a – 3)^{2} – 6a < 0, or 5a^{2} – 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**

Last one.

**Q3. **Find the set of values of ‘a’ for which the point (a, a + 1) lies in the minor segment of the circle x^{2} + y^{2} = 25 formed by the line 2x + y – 4 = 0

**Sol. **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. In case we needed to be in the major segment, we would have kept it on the same side as that of the center.

Makes sense? Now let’s do the math.

For the point to lie inside the circle, S_{1} < 0. That is a^{2} + (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.