Question

The values of $\lambda$, so that the line $3x-4y=\lambda$ touches $x^2+y^2-4x-8y-5=0$ are

• A. $-35,15$
• B. $3,-5$
• C. $35,-15$
• D. $-3,5$
• E. $20,5$

Answer 1

The correct option is A

$-35,15$

Explanation for correct option:

Given $x^2+y^2-4x-8y-5=0$

Comparing given equation with the general equation

$x^2+y^2+2gx+2fy+c=0$

Center is $C(-g,–f)=(2,4)$

radius$=\sqrt{g^2+f^2-c}$

$$\begin{gathered} =\sqrt{{\left(-2\right)}^2+{\left(-4\right)}^2+5} \\ =5 \end{gathered}$$

We know that distance of a straight line $ax+by+c=0$ from any point $\left(x_1,y_1\right)$ is $d=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}$

$$\begin{gathered} \therefore 5=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}} \\ \Rightarrow 5=\frac{\left|6-16-\lambda \right|}{\sqrt{16+9}} \\ \Rightarrow 5=\frac{\left|-\lambda -10\right|}{5} \\ \Rightarrow \frac{-\lambda -10}{5}=\pm 5 \\ \Rightarrow -\lambda -10=\pm 25 \\ \Rightarrow \lambda =\pm 25-10 \\ \Rightarrow \lambda =-35,15 \end{gathered}$$

Hence option A is correct.