The absolute difference between the squares of the radii of the two circles passing throug…

Question

The absolute difference between the squares of the radii of the two circles passing through the point $(-9,4)$ and touching the lines $x+y=3$ and $x-y=3$, is equal to \_\_\_\_\_\_\_\_ .

Answer: 768

Accepted Answer

$\because x+y=3$ and $x-y=3$ are tangents

$	herefore \,$ Both circle centre will lie on $x$-axis

$	herefore (x-a{)}^{2}+{y}^{2}={r}^{2}$

Hence centre is $C(\alpha ,0)$

$\begin{gathered} r=\sqrt{(\alpha +9{)}^{2}+16}\, \ |\frac{\alpha -3}{\sqrt{2}}|=r\, \ \sqrt{(\alpha +9{)}^{2}+16}=|\frac{\alpha -3}{\sqrt{2}}| \ \Rightarrow \, \alpha =-5-37 \ r=|\frac{-5-3}{\sqrt{2}}||\frac{-37-3}{\sqrt{2}}| \ =4\sqrt{2}20\sqrt{2} \ |{r}_{2}^{1}-{r}_{2}^{2}|=|32-800|=768 \end{gathered}$

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