Question

The range of values of $r$, for which the point $(-5+\frac{5}{\sqrt{2}},-3+\frac{r}{\sqrt{2}})$ is an interior point of the major segment of the circle $x^2+y^2=16$ cut-off by the line $x+y=2$, is

• A. $(-\infty ,5\sqrt{2})$
• B. $(4\sqrt{2},\sqrt{14},5\sqrt{2})$
• C. $(4\sqrt{2}-\sqrt{14},4\sqrt{2}+\sqrt{14})$
• D. None of the above

Answer 1

Answer: (B)

$(4\sqrt{2},\sqrt{14},5\sqrt{2})$

Since, the given point is an interior point.
then, ${(-5+\frac{r}{\sqrt{2}})}^2+{(-3+\frac{r}{\sqrt{2}})}^2-16<0$
$\Rightarrow r^2-8\sqrt{2}r+18<0$
$\Rightarrow 4\sqrt{2}-\sqrt{14}<r<4\sqrt{2}+\sqrt{14}$
The point is on the major segment.
The centre and the point are on the same sides of the line $x+y=2$.
$\therefore -5+\frac{r}{\sqrt{2}}-3+\frac{r}{\sqrt{2}-2}<0$
$\Rightarrow r<5\sqrt{2}$, so $4\sqrt{2}-\sqrt{14}<r<5\sqrt{2}$.