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The equation of the circle having radius ‘5’ and which touches the line 3x + 4y -11 = 0 at the point (1,2)

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${(x - 3)}^{2} + {(y + 1)}^{2} = 25$

${(x + 3)}^{2} + {(y + 1)}^{2} = 25$

${(x + 5)}^{2} + {(y + 5)}^{2} = 25$

${(x - 5)}^{2} + {(y + 5)}^{2} = 25$

answer is A.

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Detailed Solution

$Centre C = (1 \pm 5 \cos θ, 2 \pm 5 \sin θ)$

$(1 \pm 5, (\frac{4}{5}), 2 \pm 5 (\frac{3}{5}))$

$C = (5, 5) or C = (- 3, - 1)$

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