Question

If the circles $x^2+y^2-8x+2y+8=0and{x}^2+y^2-2y-6y+10-a^2=0$ have exactly two common tangents then

• A. $1<|a|<8$
• B. $2<|a|<8$
• C. $3<|a|<8$
• D. $4<|a|<8$

Answer 1

The correct option is B $2<|a|<8$

$S_1\equiv x^2+y^2-8x+2y+8=0$

$r_1=\sqrt{16+1-8}$

$=\sqrt{9}=3$

$S_2\equiv x^2+y^2-2x-6y+10-a^2=0$

$r_2\sqrt{1+9-(10-a^2)}$

We know the condition for two tangents that

Distance between centres $<r_1+r_2$

$\Rightarrow \sqrt{{(4-1)}^2+{(-1-3)}^2}<3+\sqrt{10-10+a^2}$

$\sqrt{3^2+4^2}$ $<3+\left|a\right|$

$\Rightarrow \pm 5<3+\left|a\right|$

When $+ve$ sign is taken or when $-ve$ sign is taken,

$5<3+\left|a\right|$ and $-5<3+\left|a\right|$

$\Rightarrow \left|a\right|>2$ and $\left|a\right|<8$

Thus, $2<\left|a\right|<8$