Question

The circle $x^2+y^2=16$ is cut by line $x+y=2$, then $(-5+\frac{r}{\sqrt{2}},-3+\frac{r}{\sqrt{2}})$ lies in the major segment if $r$ belongs to

• A. $(4\sqrt{2}-14,4\sqrt{2}+14)$
• B. $(-\infty ,\frac{5}{2})$
• C. $(4\sqrt{2}-14,5\sqrt{2})$
• D. $(0,1)$

Answer 1

The correct option is C $(4\sqrt{2}-14,5\sqrt{2})$

Given: $(-5+\frac{r}{\sqrt{2}},-3+\frac{r}{\sqrt{2}})$ lies in the major segment.

Then, $(0,0)$ and $(-5+\frac{r}{\sqrt{2}},-3+\frac{r}{\sqrt{2}})$ are along same side of $x+y=2$.

$\therefore \frac{(-5+\frac{r}{\sqrt{2}}-3+\frac{n}{\sqrt{2}}-2)}{-2}>0$

$10-\sqrt{2}r>0$

$r<5\sqrt{2}$ ----(1)

and $(-5+\frac{r}{\sqrt{2}},-3+\frac{r}{\sqrt{2}}$ lies inside circle $x^2+y^2=16$

$(-5+\frac{r}{\sqrt{1}},-3+\frac{r}{\sqrt{2}}-16<0$

$\frac{2\times r^2}{2}+34-5\sqrt{2}r-3\sqrt{2}r-16<0$

$r^2-8\sqrt{2}r+18<0$

$r+[4\sqrt{2}-4\sqrt{14},5\sqrt{2}]$----(2)

From (1) & (2)

$r+[4\sqrt{2}-\sqrt{14},5\sqrt{2}]$