Question

Two circles $x^2+y^2-2kx=0$ and $x^2+y^2-4x-8y+16=0$ touch each other externally. Then, $k$ is

• A. $4$
• B. $1$
• C. $2$
• D. $-4$

Answer 1

The correct option is C $2$

General equation of circle is $x^2+y^2+2gx+2fy+c=0$
Centre $(-g,-f)$ Radius$=\sqrt{g^2+f^2-c}$
$x^2+y^2-2kx=0$ $x^2+y^2-4x-8y+16=0$
$C_1=(k,0)C_2=(2,4)$
$R_1=\sqrt{{(-k)}^2}=k{R}_2=\sqrt{{(-2)}^2+{(-4)}^2-16}=2$
If both circles touch each other externally then $C_1{C}_2=R_1+R_2$
$\sqrt{{(k-2)}^2+{(0-4)}^2}=k+2$
${(k-2)}^2+16={(k+2)}^2$
$k^2+4-4k+16=k^2+4+4k$
$8k=16\Rightarrow k=2$