Question

If the point $(2,k)$ lies outside the circles $x^2+y^2+x-2y-14=0$ and $x^2+y^2=13$, then

• A. $k\in (-3,-2)\cup (3,4)$
• B. $k\in (-3,4)$
• C. $k\in (-\infty ,-3)\cup (4,\infty )$
• D. $k\in (-\infty ,-2)\cup (3,\infty )$

Answer 1

Answer: (C)

$k\in (-\infty ,-3)\cup (4,\infty )$

Since the point $(2,k)\equiv (x,y)$ lies outside the circle

$x^2+y^2+x-2y-14=0$

$\therefore 4+k^2+2-2k-14>0\Rightarrow k^2-2k-8>0$

$\Rightarrow (k+2)(k-4)>0\Rightarrow k\in (-\infty ,-2)\cup (4,\infty )$ $...\left(1\right)$

Also, the point $(2,k)$ lies outside the circle

$\Rightarrow x^2+y^2=13$

$\therefore 4+k^2-13>0\Rightarrow k^2-9>0$

$\Rightarrow (k-3)(k+3)>0\Rightarrow k\in (-\infty ,-3)\cup (3,\infty )$ $...\left(2\right)$

The common solution of $\left(1\right)$ and $\left(2\right)$ is given by

$k\in (-\infty ,-3)\cup (4,\infty )$