Question

If the circles $x^2+y^2+ax-6y+4=0$ and $x^2+y^2-12x+32=0$ touch externally each other and if the distance between the centres is equal to $5$, then the values of $a$ are

• A. $2,3$
• B. $-2,-3$
• C. $-4,-20$
• D. $-7,-6$
• E. $-6,-4$

Answer 1

Answer: (C)

$-4,-20$

Let $S_1\equiv x^2+y^2+ax-6y+4=0$
Its centre, $C_1\equiv (-\frac{a}{2},3)$
and $S_2\equiv x^2+y^2-12x+32=0$
Its centre, $C_2\equiv (6,0)$
Given, $C_1{C}_2=5$
$\Rightarrow (C_1{C}_2{)}^2=25$
$\Rightarrow (6+\frac{a}{2})+(0-3{)}^2=25$
$\Rightarrow 36+\frac{a^2}{4}+6a+9=25$
$\Rightarrow 144+a^2+24a+36=100$
$\Rightarrow a^2+24a+80=0$
$\Rightarrow (a+4)(a+20)=0$
Therefore, $a=-4,-20$