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Question: In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C?
- 36
- 72
- 89
- 96
- 109
Approach Solution (1)
\(x^2+y^2=62\)
\(2x^2=36\)
\(x=3\sqrt2=4.2 approx\)
Coordinates of P = (4.2, 4.2)
All points with equal x and y coordinates will lie on the green line i.e. (1, 1), (2, 2), (3, 3) and (4,4) and these 4 will be inside the circle.
Total points inside the circle below the green line in the first quadrant = 5+4+3+2 = 14
Similarly there will be 14 points on the other side of the green line in the first quadrant.
First quadrant will have a total of 4 + 14 + 14 = 32 points
All four quadrants will have 4*32 = 128 points
But 5 points on the x and y-axis are double counted
So, subtract 4 * 5 = 20 out of the total 128 to get 108 points
Now, counting the coordinates (0, 0), we have a total of 109 points.
Correct option: E
Approach Solution (2)
We have \(x^2+y^2<36\) for points to lie inside the circle
Taking integer squares we can consider 0, 1, 2, 3, 4, 5, -1, -2, -3, -4, -5
11 integers in total
For 0, 1, 2, 3, -1, -2, -3: we can have 11 possible pairs (including itself): 11 * 7 = 77
For 4, -4: we can't have 5 or -5: 2 * (11-2) = 18
For 5, -5: we can't have 5, -5, 4, -4: 2 * (11 - 4) = 14
77 + 18 + 14 = 109
Correct option: E
Approach Solution (3)
The equation of the circle is: \((x+2)^2+(y-1)^2=36\)
So, for all points inside the circle the expression becomes\( (x+2)^2+(y-1)^2<36\)
We need to find integer solutions(x,y) for this
So both (x+2) and (y-1) can range from min -5 to max 5
We need to find the no of suitable combinations
For each value of (x+2) and (y-1) we get unique values for x and y
So, let’s say x + 2 = a and y – 1 = b
\(a^2+b^2=36\)
For a = 0: b = -5 to 5...11 sets of (a, b)
For a = 1/-1: b = -5 to 5...11 * 2 = 22 sets
For a = 2/-2: b = -5 to 5...11 * 2 = 22 sets
For a = 3/-3: b = -5 to 5...11 * 2 = 22 sets (since 32+52<36)
For a = 4/-4: b = -4 to 4...9 * 2 = 18 sets (since 42+52<36 but 42+32<36)
For a = 5/-5: b = -3 to 3...7 * 2 = 14 sets (since 52+42>36 but 52+32<36)
So total no of solutions = 11 + 22 + 22 + 22 + 18 + 14 = 109
Correct option: E
“In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
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