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  • If the line touches the circle <img alt="\mathrm{x^2+y^2-4 x-8 y-5=0}" src=

If the line \mathrm{3 x-4 y-k=0} touches the circle \mathrm{x^2+y^2-4 x-8 y-5=0} at (a, b), then the positive integral value of \mathrm{\frac{k+a+b}{5}=} __________.

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Equation of circle is \mathrm{ x^2+y^2-4 x-8 y-5=0} Centre is (2,4) and radius,

\mathrm{ r=\sqrt{4+16+5}=5\, \, Line \, \, 3 x-4 y-k=0} touches the circle.

\mathrm{ \begin{aligned} & \therefore \frac{(3 \times 2)+(-4 \times 4)-k}{\sqrt{9+16}}= \pm 5 \Rightarrow k=15 \text { or }-35 \\\\ & \therefore \quad k=15 \end{aligned} }

Now equation of the tangent at (a, b) is

\mathrm{ \begin{aligned} & x a+y b-2(x+a)-4(y+b)-5=0 \\ & \text { i.e., }(a-2) x+(b-4) y-(2 a+4 b+5)=0 \end{aligned} }

If it represents the given line \mathrm{ 3 x-4 y-k=0}, then

\mathrm{ \frac{a-2}{3}=\frac{b-4}{-4}=\frac{2 a+4 b+5}{k}=\lambda }

On simplification, \mathrm{\lambda=1, a=5, b=0}

So, \mathrm{\frac{k+a+b}{5}=\frac{20}{5}=4}

Posted by

Divya Prakash Singh

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