Find the centre and the radius of the circle. (a) x2 + y2 – 4x –

1.	Find the centre and the radius of the circle. (a) x2 + y2 – 4x – y – 5 = 0 (b) 3x2 + 3y2 – 6x – 12y – 2 = 0 (c) (x – 2) (x – 4) + (y – 1) (y – 3) = 0 (d) x2 + y2 – 2x cosα – 2y sinα = 1
Answer» (a) x² + y² – 4x – y – 5 = 0 Comparing with x² + y² + 2gx + 2fy + C = 0 We get 2g = – 4, 2f = – 1 ⇒ g = -2, f = \(\frac{1}{2}\) ∴ C=(-8,-1) = (2, \(\frac{1}{2}\)) (b) Given 3x² + 3y² – 6x – 12y – 2 = 0, divide by 3 x²+ y²– 2x – 4y – \(\frac{2}{3}\) = 0 Here g = -1, f = -2, & c = \(\frac{2}{3}\) ∴Centre = (-g, -f) = (1, 2) & r = \(\sqrt{g^2 + f^2 - c}\) = \(\sqrt{(-1)^2 + (-2)^2 - (-\frac{2}{3}})\) r = \(\sqrt{1 - 4 - \frac{2}{3}}\) = \(\sqrt{\frac{17}{3}}\) units. (c) Given (x – 2) (x – 4) + (y – 1) (y – 3) = 0 ⇒ x² – 6x + 8 + y² – 4y + 3 = 0 ⇒ x² + y² – 6x – 4y + 11 = 0 Here g = -3, f= -2, C = 11 & ∴ Centre = (3, 2). And r = \(\sqrt{9 + 4 - 11}\) = √2 units (d) Given x² + y² – 2xcosα – 2ysinα – 1 = 0 g = – cosα, f = – sinα, c = -1 ∴ Centre = (cosα, sinα) & r = \(\sqrt{cos^\alpha + sin^2\alpha + 1} = \sqrt{1 + 1}\) = √2

Find the centre and the radius of the circle. (a) x2 + y2 – 4x – y – 5 = 0 (b) 3x2 + 3y2 – 6x – 12y – 2 = 0 (c) (x – 2) (x – 4) + (y – 1) (y – 3) = 0 (d) x2 + y2 – 2x cosα – 2y sinα = 1

Answer»

(a) x² + y² – 4x – y – 5 = 0

Comparing with x² + y² + 2gx + 2fy + C = 0

We get 2g = – 4, 2f = – 1 ⇒ g = -2, f = \(\frac{1}{2}\)

∴ C=(-8,-1) = (2, \(\frac{1}{2}\))

(b) Given 3x² + 3y² – 6x – 12y – 2 = 0, divide by 3

x²+ y²– 2x – 4y – \(\frac{2}{3}\) = 0

Here g = -1, f = -2, & c = \(\frac{2}{3}\)

∴Centre = (-g, -f) = (1, 2) & r = \(\sqrt{g^2 + f^2 - c}\)

= \(\sqrt{(-1)^2 + (-2)^2 - (-\frac{2}{3}})\)

r = \(\sqrt{1 - 4 - \frac{2}{3}}\) = \(\sqrt{\frac{17}{3}}\) units.

⇒ x² – 6x + 8 + y² – 4y + 3 = 0

⇒ x² + y² – 6x – 4y + 11 = 0

Here g = -3, f= -2, C = 11 &

∴ Centre = (3, 2).

And r = \(\sqrt{9 + 4 - 11}\) = √2 units

(d) Given x² + y² – 2xcosα – 2ysinα – 1 = 0

g = – cosα, f = – sinα, c = -1

∴ Centre = (cosα, sinα) & r

= \(\sqrt{cos^\alpha + sin^2\alpha + 1} = \sqrt{1 + 1}\) = √2