Question

If the curves, $x^2-6x+y^2+8=0$ and $x^2-8y+y^2+16-k=0,(k>0)$ touch each other at a point, then the largest value of $\mathrm{k}$ is:

Answer 1

Step 1: Determine the center & radius

Given, curves, $x^2-6x+y^2+8=0$ and $x^2-8y+y^2+16-k=0,(k>0)$ touch each other at a point.

Comparing curve $x^2-6x+y^2+8=0$ with $x^2+y^2+2gx+2fy+c=0$ where $\left(-g,-f\right)$ is center & $radius=\sqrt{f^2+g^2-c}$ we have:

Center$\equiv \left(3,0\right)$

$\begin{array}{rcl}\mathrm{radius}& =& \sqrt{3^2+0-8}\\ & =& 1\end{array}$

Comparing curve $x^2-8y+y^2+16-k=0,(k>0)$ with $x^2+y^2+2gx+2fy+c=0$ we get:

Center$\equiv \left(0,4\right)$

$\begin{array}{rcl}\mathrm{radius}& =& \sqrt{0+4^2+\mathrm{k}}\\ & =& \sqrt{16+\mathrm{k}}\end{array}$

Step 2: Find distance between center

Circle touch each other if ${\mathrm{C}}_1{\mathrm{C}}_2=\left|{\mathrm{r}}_1\pm {\mathrm{r}}_2\right|$
Distance between center ${\mathrm{C}}_2(0,4)$ and ${\mathrm{C}}_1(3,0)$ is either $\sqrt{\mathrm{k}}+1$ or $|\sqrt{\mathrm{k}}-1|$.

$\begin{array}{rcl}{C}_1{C}_2& =& \sqrt{3^2+4^2}\\ & =& 5\end{array}$

Step 3: Find maximum value of $k$

To find the maximum value of $k$put ${\mathrm{C}}_1{\mathrm{C}}_2=\left|{\mathrm{r}}_1\pm {\mathrm{r}}_2\right|$.

$$\begin{gathered} \Rightarrow \sqrt{\mathrm{k}}+1=5 \\ \Rightarrow \sqrt{k}=4 \\ \Rightarrow \mathrm{k}=16 \end{gathered}$$

For, $|\sqrt{\mathrm{k}}-1|=5$
$$\begin{gathered} \Rightarrow \sqrt{\mathrm{k}}-1=5 \\ \Rightarrow \sqrt{\mathrm{k}}=5+1 \\ \Rightarrow \sqrt{\mathrm{k}}=6 \\ \Rightarrow \mathrm{k}=36 \end{gathered}$$
Maximum at $k=36$.
Therefore, the largest value of $k$ is 36.