How many points with Integer x and Y co-ordinates lie within the circl

Hi

Yes, I agree with mdacosta .. If you are looking at all the integer points lying within

x^2 + y^2 < 16
Then those are 45, NOT 37.

One way to do is to rewrite the inequality like this:
y^2 < 16 - x^2

Here if x=0, then we have y^2 < 16 Or -4 < y < 4. y can take 7 integer values here (-3, -2, -1, 0, 1, 2, 3)

If x=1, then we have y^2 < 15. Here also y can take 7 integer values (-3, -2, -1, 0, 1, 2, 3)
If x=-1, then also we have y^2 < 15. Here also y can take 7 integer values (-3, -2, -1, 0, 1, 2, 3)

If x=2, then we have y^2 < 12. Here also y can take 7 integer values (-3, -2, -1, 0, 1, 2, 3)
If x=-2, then also we have y^2 < 15. Here also y can take 7 integer values (-3, -2, -1, 0, 1, 2, 3)

If x=3, then we have y^2 < 7. Here y can take 5 integer values (-2, -1, 0, 1, 2)
If x=-3, then also we have y^2 < 7. Here also y can take 5 integer values (-2, -1, 0, 1, 2)

We cannot take x as any other integer value as that would mean y^2 < 0

So total required points with integer coordinates = 7 + 7 + 7 + 7 + 7 + 5 + 5 = 45.

Yes, it should be 45 and not 37
11,12,13,21,22,23,31,32-similar points in other coordinates 8*4
01,02,03-similar points in other coordinaters 3*4
00
sum 45

When you plot the graph, the result will look as in the attachment. The radius of the circle being 4. The no of points inside the circle =3*3*4=36+(0,0)=37.

Let me know if anything more needs to be explained.

Hello VeritasPrepKarishma / mikemcgarry,
can you please give us an insight on this?

Hii

Please see my attached sketch.
Feel free to ask again if you have any more doubts.

Please recheck the OA.

gmatbusters,
exactly! the oa is wrong. thanks for your post.

I think I should post this Solution with an apology for mentioning a wrong answer before. I didn't realize it until today so thank you all forum users for reminding me to update the OA and post solutions.

I think the equation of Circle should be x^2 + y^2 = 16

and the points must satisfy the expression x^2 + y^2 ≤ 16 with an equal to sign along with less than sign.

This gives us 49 points in total as presented in the figure attached.

Please let me know in case of any discrepancies.

hii

Thanks for the Explanation, but the points on the circle itself can't be considered as points within the circle.
Or, does points within the circle include the points on the circumference ?
Bunuel, chetan2u

"Point within circle" includes

- Points on the circumference
- Points inside the circle
Both



Posted from my mobile device

This is my take on the question:

"within the circle" - implies not beyond so I would assume we are looking at points on or inside the circle

Parabola y = ax^2 + 4
ax^2 + 4 is nothing but a quadratic and since a is positive, I know it is an upward open parabola.

y = ax^2 would be drawn at the origin but since we have +4 too, I will move it 4 units up.
The circle has origin as the centre and a single point intersection with the parabola so it must intersect at the point (0, 4) only, the lowest point of the parabola. If it intersects at any point higher than (0, 4), it will intersect at two points.

So my unique circle is ready. I will focus on one quadrant and multiply that by 4.
The line y = x will divide the 1st quadrant into two equal halves. It will cut the circle at \((2\sqrt{2}, 2\sqrt{2})\) which is about (2.8, 2.8), So we know that the circle will be just shy of the (3, 3) point. So we can just draw it out and find the number of points which are on/inside it. Just draw lines across x and y co-ordinates which are integers. The intersection of these lines will be those points which will have both co-ordinates as integers.

We see that there will be 12 such points.
So we will have a total of 12*4 + 1 (the origin) = 49 points

Thats a new terminology i learned today...
Thanks...

Within = inside or on circumference.

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