**Example 1. **Find the equation of the circle, whose diameter’s end points are (1, 3) and (4, 2).

**Sol. **Pretty simple. Using the form we derived in the previous lesson, the required equation is **(x – 1)(x – 4) + (y – 3)(y – 2) = 0**, or **x ^{2} + y^{2} – 5x – 5y + 10 = 0.**

**Example 2. **Find the equation of the circle passing through the points (0, 0), (a, 0) and (0, b).

**Sol. **I’ve covered a similar example in a previous lesson (see Example 9). But now we have a better method to obtain the equation.

Observe that the points (a, 0) and (0, b) lie on the X and the Y axes respectively. That means angle AOB = 90°, which implies that AB must be the diameter of the circle.

Therefore the required equation is **(x – a)(x – 0) + (y – 0)(y – b) = 0**, or **x ^{2} + y^{2} – ax – by = 0.**

Note that the center of the circle is (a/2, b/2) (midpoint of AB, the diameter), and the radius is \( \sqrt{a^2+b^2}/2 \) (half of AB, the diameter)

**Example 3. **Find the equation of the circle, whose chord with end-points as (1, 2) and (0, 4) subtends an angle of 45° at its circumference.

**Solution.** This equation type was also covered in the previous lesson.

I’ll use that equation directly: (x – 1)(x – 0) + (y – 2)(y – 4) = ± cot45° [(y – 2)(x – 0) – (y – 4)(x – 1)].

As I mentioned earlier, there’ll be two different circles satisfying the given conditions: **x ^{2} + y^{2} – 3x – 7y + 12 = 0** and

**x**

^{2}+ y^{2}+ x – 5y + 4 = 0.

And that’s it for this lesson. See you in the next with the parametric equation of the circle.