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)

Given, $x^2+y^2=16x$
$\Rightarrow x^2-16x+y^2=0$
$\Rightarrow x^2-2\left(x\right)8+{\left(8\right)}^2-{\left(8\right)}^2+y^2=0$
$\Rightarrow {\left(x-8\right)}^2+y^2=64...\left(i\right)$
$\therefore$ Centre = (8, 0) and radius = 8
Since, the straight line $3x+4y=k$ touches the circle $x^2+y^2=16$, therefore the length of perpendicular from the centre (8, 0) to the straight line $3x+4y-k=0$ is equal to the radius of the circle.
i.e., $8=\left|\frac{3\left(8\right)+4\left(0\right)-k}{\sqrt{9+16}}\right|$
$\Rightarrow 8=\left|\frac{24-k}{\sqrt{25}}\right|$
$\Rightarrow 8=\frac{|24-k|}{5}$
$\Rightarrow \left|24-k\right|=40$
$\Rightarrow 24-k=\pm 40$
Taking positive sign, we get
$24-k=40$
$\Rightarrow -k=-40-24=16$
$\Rightarrow k=-16$
Taking negative sign, we get
$24-k=-40$
$\Rightarrow -k=-40-24=-64$
$\Rightarrow k=64$
$\therefore k=-16,64$