Equation of circle touching the line |x-2|+|y-3|=4 will be

$(x - 2)^{2} + (y - 3)^{2} = 12$

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$(x - 2)^{2} + (y - 3)^{2} = 4$

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$(x - 2)^{2} + (y - 3)^{2} = 10$

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$(x - 2)^{2} + (y - 3)^{2} = 8$

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Solution

The correct option is D
$(x - 2)^{2} + (y - 3)^{2} = 8$

Perpendicular distance from centre to tangent = radius

$r = \frac{| 2 + 3 - 9 |}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2}$ Equation of circle is $(x - 2)^{2} + (y - 3)^{2} = 8$

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