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