Question

If the circle x² + y² – kx – 12y + 4 = 0 touches x-axis, then k = __________.

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{equation}\mathrm{is} \\ {x}^2+y^2-kx-12y+4=0...\left(1\right) \\ \mathrm{and}\mathrm{general}\mathrm{equation}\mathrm{is}{\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2 \\ \mathrm{i}.\mathrm{e}.x^2+y^2-2ax-2by+a^2+b^2=r^2 \\ \mathrm{i}.\mathrm{e}.x^2+y^2-2ax-2yb+a^2+b^2-r^2=0...\left(2\right) \\ \left.\begin{array}{r}\mathrm{i}.\mathrm{e}.a^2+b^2-r^2=4\\ k=2ai.e.a=\frac{k}{2}\end{array}\right\}\mathrm{as}\mathrm{comparing}\left(1\right)\mathrm{and}\left(2\right) \\ b=+6 \\ \mathrm{also},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{circle}\mathrm{touches}x-\mathrm{axis} \\ \mathrm{i}.\mathrm{e}.a^2=ri.e.{\left(\frac{k}{2}\right)}^2=r \\ \mathrm{and}r+36-r^2=4 \\ \mathrm{i}.\mathrm{e}{r}^2-r-32 \\ \mathrm{i}.\mathrm{e}r=4 \\ \mathrm{i}.\mathrm{e}.\frac{k^2}{4}=r \\ \mathrm{i}.\mathrm{e}.k^2=16 \\ \mathrm{i}.\mathrm{e}.k=\pm 4 \end{gathered}$$