**Q1. **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**

**Q2. **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 Q9). 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)

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