Question

If the point (2, k) lies outside the circles x² + y² + x − 2y − 14 = 0 and x² + y² = 13 then k lies in the interval
(a) (−3, −2) ∪ (3, 4)
(b) −3, 4
(c) (−∞, −3) ∪ (4, ∞)
(d) (−∞, −2) ∪ (3, ∞)

Answer 1

(c) (−∞, −3) ∪ (4, ∞)

The given equations of the circles are x² + y² + x − 2y − 14 = 0 and x² + y² = 13.

Since (2, k) lies outside the given circles, we have:

$4+k^2+2-2k-14>0$ and $4+k^2>13$
$\Rightarrow k^2-2k-8>0$ and $k^2>9$
$\Rightarrow \left(k-4\right)\left(k+2\right)>0$ and $k^2>9$
$\Rightarrow k>4\mathrm{or}\mathrm{k}<-2$ and $k>3\mathrm{or}k<-3$
$\Rightarrow k>4\mathrm{and}\mathrm{k}<-3$
$\Rightarrow k\in \left(-\infty ,-3\right)\cup \left(4,\infty \right)$