Question

The range of values of $r$, for which the point $(-5+\frac{r}{\sqrt{2}},-3+\frac{r}{\sqrt{2}})$ 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

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


Position of origin w.r.t line $x+y-2\Rightarrow 0-0-2=-2<0$
therefore, major segment of the circle will lie left side of line
Point $P(-5+\frac{r}{\sqrt{2}},-3+\frac{r}{\sqrt{2}})$ w.r.t. line is
$\Rightarrow -5+\frac{r}{\sqrt{2}}+(-3)+\frac{r}{\sqrt{2}}-2<0$
$\Rightarrow r<5\sqrt{2}...\left(1\right)$
As we know, point $P$ should lie in the left side of the line but inside the circle also
$\therefore OP<r$
$\Rightarrow \sqrt{{(-5+\frac{r}{\sqrt{2}})}^2+{(-3+\frac{r}{\sqrt{2}})}^2}<4$
$\Rightarrow r^2-8\sqrt{2}r+18<0$
$\Rightarrow r\in (4\sqrt{2}-14,4\sqrt{2}+14)...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow r\in (4\sqrt{2}-14,5\sqrt{2})$