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If the line \mathrm{3 x-4 y-\lambda=0} touches the circle \mathrm{x^2+y^2-4 x-8 y-5=0} at (a, b), then \mathrm{\lambda+a+b=}

 

Option: 1

-28


Option: 2

20


Option: 3

+26


Option: 4

-8


Answers (1)

best_answer

As \mathrm{3 x-4 y-\lambda=0} is a tangent to the circle

\mathrm{x^2+y^2-4 x-8 y-5=0}  with centre C(2, 4) and radius

\mathrm{r=\sqrt{4+16+5}=5}, the condition is that

the perpendicular distance from C on this line is equal to the radius

\mathrm{\begin{aligned} & \Rightarrow \quad \frac{3(2)-4(4)-\lambda}{ \pm \sqrt{9+16}}=5 \\ & \Rightarrow \lambda=15,-35 \end{aligned}}

Tangent at (a, b) to the given circle is

      \mathrm{x a+y b-2(x+a)-4(y+b)-5=0}

Or \mathrm{(a-2) x+(b-4) y-(2 a+4 b+5)=0 \text {, }} -------------(1)

If it  is the same as \mathrm{3 x-4 y-\lambda=0,} -----------(2)

the conditions are \mathrm{(a-2) x+(b-4) y-(2 a+4 b+5)=0 \text {, }}

\mathrm{\begin{aligned} & \therefore \quad a=3 k+2, b=4-4 k, 2 a+4 b+5=\lambda k \\ & \Rightarrow 2(3 k+2)+4(4-4 k)+5 \quad=\lambda k \text {, But } \lambda=15,-35 \\ & \Rightarrow \quad-10 k+25=\left[\begin{array}{c} 15 k \\ -35 k \end{array}\right. \\ & \end{aligned}}------(4)

\mathrm{\begin{aligned} & \text { Or }-25 k=-25 \Rightarrow k=1 \Rightarrow a=5, b=0, \\ & \lambda=10+0+5=15 \\ & \Rightarrow \lambda+a+b=15+15+0=20 \end{aligned}}

Similarly, from (4)

\mathrm{\begin{aligned} & k=-1 \Rightarrow a=-1, b=8, \lambda=-35 \\ & \therefore \quad \lambda+a+b=-35-1+8=-28\end{aligned}}

 

 

 

Posted by

Irshad Anwar

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