Question

The line $4y-3x+\lambda -0$ touches the circle $x^2+y^2-4x-8y-5=0$, the value of $\lambda$ is?

• A. $29$
• B. $10$
• C. $-35$
• D. None of these

Answer 1

The correct option is C $-35$

If line $4y-3x+\lambda =0$ touches the circle $x^2+y^2-4x-8y-5=0$ then distance of its centre from the line is equal to its radius.

Centre is $(-2,-4)$

radius $=\sqrt{4+16+5}=\sqrt{25}=5$ units

$\therefore$ $5=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}$

$5=\frac{|6-16+\lambda |}{\sqrt{16+9}}\Rightarrow 5=\frac{|-10+\lambda |}{5}$

$\Rightarrow \frac{\lambda -10}{5}=\pm 5\Rightarrow \lambda -10=25$

$\Rightarrow \lambda =10+10=35$

$\frac{\lambda -10}{5}=-5\Rightarrow \lambda -10=-25\Rightarrow \lambda =-15$

$\therefore$ $\lambda =-15,25$