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A straight line $\mathrm{x}=\mathrm{y}+2$ touches the circle $4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{r}^{2}$. The value of r is $\quad[2015-I I]$
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Verified Answer
The correct answer is:
$2 \sqrt{2}$
$\because 4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{r}^{2}$
$\Rightarrow x^{2}+y^{2}=\left(\frac{r}{2}\right)^{2}$
Center $(0,0)$ and radius $\frac{\mathrm{r}}{2}$ Eq. of line is $; \mathrm{x}-\mathrm{y}-2=0$ $\because$ Line touches the circle.
$$
\therefore \frac{\mathrm{r}}{2}=\frac{|0-0-2|}{\sqrt{(1)^{2}+(-1)^{2}}}
$$
$\mathrm{r}=2 \sqrt{2}$
$\Rightarrow x^{2}+y^{2}=\left(\frac{r}{2}\right)^{2}$
Center $(0,0)$ and radius $\frac{\mathrm{r}}{2}$ Eq. of line is $; \mathrm{x}-\mathrm{y}-2=0$ $\because$ Line touches the circle.
$$
\therefore \frac{\mathrm{r}}{2}=\frac{|0-0-2|}{\sqrt{(1)^{2}+(-1)^{2}}}
$$
$\mathrm{r}=2 \sqrt{2}$
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