Circle Ques 17
- The straight line $2 x-3 y=1$ divide the circular region $x^{2}+y^{2} \leq 6$ into two parts. If $S=2, \frac{3}{4}, \frac{5}{2}, \frac{3}{4}, \frac{1}{4},-\frac{1}{4}, \frac{1}{8}, \frac{1}{4}$, then the number of point (s) in $S$ lying inside the smaller part is … .
Paragraph Based Questions
Let $S$ be the circle in the $X Y$-plane defined by the equation
$ x^{2}+y^{2}=4 $
(2018 Adv.)
(There are two questions based on above Paragraph, the question given below is one of them)
Show Answer
Answer:
Correct Answer: 17.(2)
Solution:
Formula:
Position of a Point with Respect to a Circle
- $x^{2}+y^{2} \leq 6$ and $2 x-3 y=1$ is shown as a line intersecting a circle
For the point to lie in the shade part, the origin and the point lie on opposite sides of the straight line $L$.
$\therefore$ For any point in shaded part $L>0$ and for any point inside the circle $S<0$.
Now, for $(2, \frac{3}{4}) \quad L: 2 x-3 y-1$
$ L: 4-\frac{9}{4}-1=\frac{3}{4}>0 $
and $S: x^{2}+y^{2}-6, S: 4+\frac{9}{16}-6<0$
$\Rightarrow \quad (2, \frac{3}{4})$ lies in shaded region.
For $(\frac{5}{2}, \frac{3}{4}), L: 5-9-1<0$ [ignore]
For $(\frac{1}{4},-\frac{1}{4}), L: \frac{1}{2}+\frac{3}{4}-1>0$
$\therefore \quad (\frac{1}{4},-\frac{1}{4})$ lies in the shaded part.
For $(\frac{1}{8}, \frac{1}{4}), L: \frac{1}{4}-\frac{3}{4}-1<0$ [neglect]
$\Rightarrow$ Only 2 points lie in the shaded part.
BETA
