The equation of circle which touches the line 5x + 12 y =1 and -Turito

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Maths-

General

Easy

Question

The equation of circle which touches the line 5x + 12 y =1 and which has its centre at (3, 4) is:

$(x - 3)^{2} + (y - 4)^{2} = {(\frac{62}{11})}^{2}$
$(x - 3)^{2} + (y - 4)^{2} = {(\frac{62}{17})}^{2}$
$(x - 3)^{2} + (y - 4)^{2} = {(\frac{62}{13})}^{2}$
none of these

The correct answer is: $left parenthesis x minus 3 right parenthesis to the power of 2 end exponent plus left parenthesis y minus 4 right parenthesis to the power of 2 end exponent equals open parentheses fraction numerator 62 over denominator 13 end fraction close parentheses to the power of 2 end exponent$

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