Question

Find the number of points with integral coordinates that are interior to the circle $x^{2}+y^{2}=16$.

Answer 1

Hint: In this question it is given that we have to find the number of points with integral coordinates that are interior to the circle $x^{2}+y^{2}=16$. So to find the solution we first consider the interior region of the circle i.e, $x^{2}+y^{2}<16$ 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 $x^{2}+y^{2}=16$ we must find the point which satisfies $x^{2}+y^{2}<16$.
So from here we can clearly say that the range of x and y is,
$$y\in \left[ -3,3\right] $$, $$x\in \left[ -3,3\right] $$.
This include coordinates (0,1),(0,2),(0,3) also, (1,0),(2,0),(3,0)
Along with their negatives (0,−1),(0,−2),(0,-3) and (−1,0),(−2,0),(0,-3).
We have now ,3+3+3+3=12 points.
Also, points whose sum of the square of x and y less than 16.
(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2).
Total 8 points which lie in the first quadrant.
Now including the points in all quadrants 8×4=32.
Total points 12+32=44, also including the origin (0,0)
We have 44+1= 45 points.
So we can say that the total number of integral points is 45.

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.