Question
The range of values of $$r$$ for which the point $$\left( { - 5 + \frac{r}{{\sqrt 2 }},\, - 3 + \frac{r}{{\sqrt 2 }}} \right)$$ 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.
$$\left( { - \infty ,\,5\sqrt 2 } \right)$$
B.
$$\left( {4\sqrt 2 - \sqrt {14} ,\,5\sqrt 2 } \right)$$
C.
$$\left( {4\sqrt 2 - \sqrt {14} ,\,4\sqrt 2 + \sqrt {14} } \right)$$
D.
none of these
Answer :
$$\left( {4\sqrt 2 - \sqrt {14} ,\,5\sqrt 2 } \right)$$
Solution :
The given point is an interior point
$$\eqalign{
& \Rightarrow {\left( { - 5 + \frac{r}{{\sqrt 2 }}} \right)^2} + {\left( { - 3 + \frac{r}{{\sqrt 2 }}} \right)^2} - 16 < 0 \cr
& \Rightarrow {r^2} - 8\sqrt 2 r + 18 < 0 \cr
& \Rightarrow 4\sqrt 2 - \sqrt {14} < r < 4\sqrt 2 + \sqrt {14} \cr} $$
The point is on the major segment $$ \Rightarrow $$ the centre and the point are on the same side of the line $$x + y = 2.$$
$$\eqalign{
& \therefore \,\, - 5 + \frac{r}{{\sqrt 2 }} - 3 + \frac{r}{{\sqrt 2 }} - 2 < 0 \cr
& \Rightarrow \,r < 5\sqrt 2 {\text{ So, }}4\sqrt 2 - \sqrt {14} < r < 5\sqrt 2 \cr} $$