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)

$4$

Given equation is

$x^2+y^2-kx-12y+4=0$ (ii)

We know that

General equation to circle is

${\left(x-a\right)}^2+{\left(x-b\right)}^2=r^2$

$x^2+y^2+2ax+2by+\left(a^2+b^2-r^2\right)=\mathrm{0..............}\left(i\right)$

Now,

Compasing equation (i) and (ii) we get

$a=\frac{-R}{2}$

$b=-6$

$r=4$

$\because$ the circle touches $x$ axis.

$\therefore$ $a^2=r$

$\begin{array}{c}{\left(-\frac{R}{2}\right)}^2=4\\ \frac{R^2}{4}=4\\ R^2=16\\ R=\sqrt{16}=4\end{array}$