Question

The range of value of a which the point $(a,4)$ is outside the circle $x^2+y^2+10x=0$ and $x^2+y^2-12x-20=0$

Answer 1

Consider the given equation of circle

$x^2+y^2+10x=\mathrm{0....}\left(1\right)$

$x^2+y^2-12x-20=\mathrm{0.......}\left(2\right)$

point $(a,4)$ outside the circle

equation (1)

$a^2+y^2+10\times a=0$

$\Rightarrow$ $a^2+10a+16=0$

$\Rightarrow$ $a^2+8a+2a+16=0$

$\Rightarrow$ $a(a+8)+2(a+8)=0$

$\Rightarrow$ $(a+8)(a+2)=0$

$\Rightarrow$ $a=-2,-8<0$ (outside the circle)

equation (2)

$a^2+y^2-12a-20=0$

$\Rightarrow$ $a^2+16-12a-20=0$

$\Rightarrow$ $a^2-12a-4=0$

Using quadratic formula

$\Rightarrow$ $a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=\frac{-(-12)\pm \sqrt{144+16}}{2}$

$a=\frac{12\pm \sqrt{150}}{2}$

Hence this is the answer