Question

If the straight line $3x+4y=k$ touches the circle $x^2+y^2=16x$, then the value of $k$ is

• A. $16,64$
• B. $-16,-64$
• C. $-16,64$
• D. $16,-64$

Answer 1

Answer: (C)

$-16,64$

The straight line $3x+4y-k=0$ touches the circle $x^2-16x+y^2=0$

The perpendicular distance between the line and the centre of the circle equals the radius.

The centre of the circle is $(8,0)$ and the radius is $\sqrt{64+0-0}=8$

$\Rightarrow \left|\frac{24-k}{\sqrt{3^2+4^2}}\right|=8$

$\Rightarrow |24-k|=40$

$\Rightarrow 24-k=\pm 40$

$\Rightarrow k=-16,64$