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, $x^{2} + y^{2} = 16x$
$\Rightarrow x^{2} -16x + y^{2} = 0$
$\Rightarrow x ^{2}- 2\left(x\right) 8 + \left(8\right)^{2} - \left(8\right)^{2} + y^{2} = 0$
$\Rightarrow \left(x-8\right)^{2}+y^{2} =64 \quad...\left(i\right)$
$\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., $8 = \left|\frac{3\left(8\right)+4\left(0\right)-k}{\sqrt{9+16}}\right|$
$\Rightarrow 8 = \left|\frac{24-k}{\sqrt{25}}\right|$
$\Rightarrow 8 = \frac{\left|24 - k\right|}{5}$
$\Rightarrow \left|24-k\right| = 40$
$\Rightarrow 24-k = \pm 40$
Taking positive sign, we get
$24 -k =40$
$\Rightarrow - k = - 40 - 24 = 16$
$\Rightarrow k = -16$
Taking negative sign, we get
$ 24-k=-40$
$\Rightarrow -k =-40-24=-64$
$\Rightarrow k =64$
$\therefore k =-16,\,64$