Question

If the total number of integral points in the interior circle $x^{2}+y^{2}=9$ are equal to 4n+1, then n is equal to
A. 1
B. 6
C. 12
D. 25

Answer 1

Hint: In this question it is given that the total number of integral points in the interior circle $x^{2}+y^{2}=9$ are equal to 4n+1, so we have to find the value of n. So to find the solution we first consider the interior region of the circle i.e, $x^{2}+y^{2}<9$ and after that we can find the integral points (x,y) which satisfies the inequation.

Complete step-by-step answer:
For integral points within the circle $x^{2}+y^{2}=9$ we must find the point such that $x^{2}+y^{2}<9$.
So from here we can clearly say that the range of x and y is,
$y\in \left[ -2,2\right]$, $x\in \left[ -2,2\right]$.
This include coordinates (0,1),(0,2) also, (1,0),(2,0)
Along with their negatives (0,−1),(0,−2) and (−1,0),(−2,0).
We have now ,2+2+2+2=8 points.
Also, points whose sum of the square of x and y less than 9.
(1,1),(1,2),(2,1),(2,2),
Total 4 points which lie in the first quadrant.
Now including the points in all quadrants 4×4=16.
Total points 8+16=24, also including the origin (0,0)
We have 24+1= 25 points.
So 25 can be written as $\left( 4\times 6+1\right)$
So n=6.
Hence, the correct option is option B.

Note: In this question it has been asked that you have to find only the integral points, that's why we have taken only the integer values of x and y in (x,y). Also you have to take all the points from each quadrant.