Question

If the circles $x^2+y^2=9$ and $x^2+y^2-2\alpha x+2y+1=0$ touch each other internally, then $\alpha =$

• A. $\pm \frac{4}{3}$
• B. $1$
• C. $\frac{4}{3}$
• D. $\frac{-4}{3}$

Answer 1

The correct option is C $\frac{4}{3}$

Circle 1: $x^2+y^2=3^2$

radius is $r_1=3$ and centre is $(0,0)$

Circle 2: $x^2+y^2-2\alpha x+{\left(\alpha \right)}^2+2\cdot 1\cdot y+1-{\left(\alpha \right)}^2+1-1=0$

${(x-\alpha )}^2+{(y+1)}^2-1-{\left(\alpha \right)}^2+1=0$

${(x-\alpha )}^2+{(y+1)}^2={\left(\alpha \right)}^2$

radius is $r_2=\alpha$ and centre is $(\alpha ,-1)$

Distance between 2 centres is diff between their radii

$r_2-r_1=C_1{C}_2$

$C_1{C}_2=\sqrt{{(\alpha -0)}^2+{(-1-0)}^2}=\sqrt{{\alpha }^2+1^2}$

$\Rightarrow \alpha -3=\sqrt{{\alpha }^2+1^2}$

$\Rightarrow {(\alpha -3)}^2={\alpha }^2+1^2\Rightarrow {\alpha }^2+9-6\alpha ={\alpha }^2+1\Rightarrow 6\alpha =8\Rightarrow \alpha =\frac{4}{3}$