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 k lies in the interval

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

Answer 1

The correct option is C

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

The given equations of the circles are

$x^2+y^2+x-2y-14=0$ and $x^2+y^2=13$

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 $k^2>9$

$\Rightarrow \left(k4\right)(k+2)>0$ and $k^2>9$

$\Rightarrow k>4$ or $k<-2$ and $k>3$ or $k<-3$

$\Rightarrow k>4$ and $k<-3$

$\Rightarrow kϵ(-\infty ,3)\cup (4,\infty )$