Question

If the circles $x^2+y^2-2x-4y=0$ and $x^2+y^2-8y-k=0$ touch each other internally, then the value of $k$ is

• A. $-16$
• B. $4$
• C. $3$
• D. $-2$

Answer 1

The correct option is B $4$

Given circles are
$S_1:x^2+y^2-2x-4y=0$ and $S_2:x^2+y^2-8y-k=0$
Now, the centre and radius of the circles are
$C_1=(1,2),r_1=\sqrt{5}{C}_2=(0,4),r_2=\sqrt{16+k}$
For the square root to be defined,
$16+k>0\Rightarrow k>-16$

Now, $C_1{C}_2=\sqrt{1^2+2^2}=\sqrt{5}$
As both the circles touch internally, so
$C_1{C}_2=|r_1-r_2|\Rightarrow \sqrt{5}=|\sqrt{5}-\sqrt{16+k}|\Rightarrow \sqrt{5}-\sqrt{16+k}=\pm \sqrt{5}\Rightarrow -\sqrt{16+k}=-\sqrt{5}\pm \sqrt{5}\Rightarrow -\sqrt{16+k}=-2\sqrt{5},0\Rightarrow k=4,-16$
But $k>-16$
$\therefore k=4$