Question

If circle $x^2+y^2-kx-12y+4=0$ touches $x-$axis , then $k=$`

• A. $\sqrt{12}$
• B. $12$
• C. $4$
• D. $16$

Answer 1

Answer: (C)

$4$

$x^2+y^2-kx-12y+4=0$ touches the $X-$axis

if the $Y-$co-ordinate of the centre of the circle is equal to the radius $[$centre of the circle is $(\frac{k}{2},6)$$]$

i.e. $6=\sqrt{\frac{k^2}{4}+36-4}$

$\Rightarrow 36=\frac{k^2}{4}+36-4$

$\Rightarrow k^2=4\times 4$

$\Rightarrow k=4$

Hence, this is the answer.