The centre will be (–g, –f) or (2, –3), and the radius will be \( \sqrt{g^2+f^2 – c} = \sqrt{2^2+{(-3)}^2 – (-3)} \) = 4

Example 7 Find the equation of the circle which touches the line 3x + 4y = 5 and has centre (2, 1).

Solution We’ve been given the centre of the circle, but not the radius. Instead, we know that the circle touches a line.

What does that tell us? Recall, from your earlier classes, that the radius of a circle is perpendicular to the tangent. In other words, the perpendicular distance from the centre to a tangent will be equal to the radius.

To find the radius, all we have to do is find the perpendicular distance of the line from the given point (i.e. the centre). Using the formula we derived earlier (distance of a point from a line), the required distance is |3(2) + 4(1) – 5|/\( \sqrt{3^2 + 4^2} \) = 1.

Therefore the required equation is (x – 2)2 + (y – 1)2 = 1

Example 8 Find the equation of circle lying in the first quadrant, which touches both the axes and has radius 3.

Solution This one’s slightly different. We’ve been given the radius of the circle, but not the centre. Instead, we know that the circle touches both the axes, and lies in the first quadrant. This is what the circle would look like.

Since the circle touches the axes, the distance of the centre from both the axes must be equal to the radius (something similar to the previous example). And, the distances from the axes are nothing but the coordinates of the centre.

Therefore the coordinates of the centre must be (3, 3). And, the required equation will be (x – 3)2 + (y – 3)2 = 32

We have a special equation – the equation of a circle touching both the axes and lying in the first (or the third) quadrant, whose radius is ‘a’, will be (x – a)2 + (y – a)2 = a2

(Also find the equation of those which lie in the 2nd or the 4th quadrant)

Example 9 Find the equation of the circle which passes through the points (0, 0), (4, 0) and (0, 3).

Solution We’ll have better and interesting ways to find equations of circles passing through three points, but I’ll cover a basic method for now.

We’ll start with the general equation, which has three unknowns: g, f and c. The values of these unknowns will be found out by forming three equations in these unknowns, obtained by substituting the given coordinates in the circle’s equation. Let’s begin.