Question

Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant.

Answer 1

Let a be the radius of the circle.

Centre of the circle = (– a, – a)

Distance of the line 3x – 4y + 8 = 0

From the centre = Radius of the circle

`|(-3a + 4a + 8)/sqrt((3)^2 + (-4)^2)|` = a

⇒ `|(a + 8)/5|` = a

⇒`+-((a + 8)/5)` = a

⇒ `(a + 8)/5` = a and `-((a + 8)/5)` = a

⇒ a = 5a – 8

⇒ 5a – a = 8

⇒ 4a = 8

⇒ a = 2

And `(a + 8)/2` = – a

⇒ a + 8 = – 5a

⇒ 6a = – 8

⇒ a = `- 4/3`

∴ a = 2 and a ≠ `-4/3`

∴ The equation of the circle is (x + 2)² + (y + 2)² = (2)²

⇒ x² + 4x + 4 + y² + 4y + 4 = 4

⇒ x² + y² + 4x + 4y + 4 = 0

Hence, the required equation of the circle

x² + y² + 4x + 4y + 4 = 0.