The equation of a circle which touches both axes and the line 3x - 4y + 8 =0 and whose centre lies in the third quadrant is

x^{2} + y^{2} - 4 x + 4 y - 4 = 0

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x^{2} + y^{2} - 4 x + 4 y + 4 = 0

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x^{2} + y^{2} + 4 x + 4 y + 4 = 0

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x^{2} + y^{2} - 4 x - 4 y - 4 = 0

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Solution

The correct option is C $x^{2} + y^{2} + 4 x + 4 y + 4 = 0$
The equation of circle in third quadrant touching the coordinate axes with centre (-a, -a) and radius ‘a’ is $x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$ and we know $∣ ∣ ∣ \frac{3 (- a) - 4 (- a) + 8}{\sqrt{9 + 16}} ∣ ∣ ∣ = a \Rightarrow a = 2.$
Hence the required equation is $x^{2} + y^{2} + 4 x + 4 y + 4 = 0.$

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