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The length of intercept on the straight line  \mathrm{5 x-3 y+2=0}  by the circle  \mathrm{x^2+y^2-4 x-8 y+11=0}  is  \mathrm{3 k} , then value of k is

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

Equation of circle is  \mathrm{ x^2+y^2-4 x-8 y+11=0}
\mathrm{ \Rightarrow(x-2)^2+(y-4)^2=9 }
\mathrm{ \therefore \quad }  Centre is C(2,4) and radius is R=3
Let CM be the \mathrm{ \perp }  from C on the line   \mathrm{ 5 x-3 y+2=0 }
Then  \mathrm{ C M=\frac{|5(2)-3(4)+2|}{\sqrt{5^2+(-3)^2}}=0 }  which means the line passes through the centre of the circle.
\mathrm{ \therefore }  Intercepted length is diameter of circle and the diameter of circle is 6 .
\mathrm{ \therefore \quad 3 k=6 \Rightarrow k=2 }

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Ajit Kumar Dubey

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