This lesson will cover a few examples to illustrate the concepts covered in the previous lesson.

**Q1. **Determine whether the given line intersects the given circle at two distinct points, touch the circle or does not intersect the circle at any point:

(i) L: 3x + 4y = 10; C: x^{2} + y^{2} = 9

(ii) L: 5x + 12y = 9; C: x^{2} + y^{2} – 4x – 2y + 4 = 0

(iii) L: x = 3; C: x^{2} + y^{2} + 4x + 6y – 3 = 0

**Sol. **The problems are quite simple, as they are directly based on what was previously discussed.

(i) I’ll illustrate the first method in this part, i.e. forming the quadratic and using its discriminant. The next two will be solved by using the second method. Let’s begin.

Substituting the value of ‘y’ from the line into the circle’s equation, we get something like x^{2} + (10 – 3x)^{2}/16 = 9, or 25x^{2} – 60x – 44 = 0. The discriminant of the equation equals 8000 which is positive. Hence, the line will intersect the circle at two distinct points.

(ii) In this part, we’ll find the distance of the given line from the circle’s center (2, 1) and compare it with its radius, which is 1.

Now the distance of the line from the point (2, 1) is given by |5(2) + 12 – 9|/13 which equals 1, i.e. equal to the radius.

Therefore the given line will touch the circle. Can you find the point where the line touches the circle?

(iii) Again, we’ll find the distance between the given line x = 3, and the center (–2, –3), which equals 5. The radius of the circle is 4, which is smaller than the calculated distance. Hence the line will not intersect or touch the circle.

**Q2. **Consider the circle x^{2} + y^{2} – 2x + 4y = 0. Find the value(s) of ‘a’ for which the line x + 2y = c

(i) touches the circle

(ii) intersects the circle at two distinct points

(iii) does not intersect the circle at any point

**Sol. **This one is quite similar to the previous one, except that the position of the line is already given, and we are to determine values of a parameter, ‘c’.

First, let’s find the distance of the given line from the center C (1, –2): CP = |3 + c|/\(\sqrt{5}\), and also the radius of the circle: a = \(\sqrt{5}\). Next, we’ll apply the conditions.

(i) In this case, CP = a, or |c + 3| = 5. Therefore c = 2 or –8.

(ii) Here, CP < a, or |c + 3| < 5. Therefore –8 < c < 2

(iii) In this case, CP > a, or |c + 3| > 5. Therefore the values of c will be c < –8 or c > 2

And that will be all for this lesson. Hope you’ve a better idea of the concept covered. The next lesson will talk about length of a chord, cut by a line on a circle.