Question

Number of integral points lying within the circle $x^2+y^2=36$ is

• A. $113$
• B. $109$
• C. $130$
• D. $110$

Answer 1

Answer: (A)

$113$

For integral points within circle $x^2+y^2=36$ we must find the point such that $x^2+y^2\le 36$

This include coordinates $(0,1),(0,2)....(0,6)$ also, $(1,0),(2,0)....(6,0)$

Along with their negatives $(0,-1),(0,-2)....(0,-6)$ and $(-1,0),(-2,0)....(-6,0)$

We have now , $12\times 2=24points$

Also, points whose sum of the square of $x$ and $y$ $<36$

$(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(4,4),(5,1),(5,2),(5,3)$

Total $22$ points

Including the points in all quadrants $22\times 4=88$

Total points $88+24=112$, also including the origin $(0,0)$

We have $112+1=113points$