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The equation of a circle in the x-y coordinate plane is x^2

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by Max@Math Revolution » Thu Nov 29, 2018 1:36 am

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[Math Revolution GMAT math practice question]

The equation of a circle in the x-y coordinate plane is x^2+y^2=25. How many points of the form (a,b), where a and b are integers, lie on this circle?

A. 4
B. 6
C. 8
D. 10
E. 12
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Source: — Problem Solving |

by GMATGuruNY » Thu Nov 29, 2018 4:28 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

The equation of a circle in the x-y coordinate plane is x^2+y^2=25. How many points of the form (a,b), where a and b are integers, lie on this circle?

A. 4
B. 6
C. 8
D. 10
E. 12
The equation for a circle with its center at the origin and a radius of r is as follows:
x² + y² = r².
Thus, x² + y² = 25 is a circle with its center at the origin and a radius of 5.

A point on this circle will have integer coordinates in three cases:
Case 1: The radius is horizontal
Case 2: The radius is vertical
Case 3: The radius could serve as the hypotenuse of a 3-4-5 triangle

Coordinate pair options for Case 1:
(5, 0)
(-5, 0)
Total ways = 2

Coordinate pair options for Case 2:
(0, 5)
(0, -5)
Total ways = 2

Coordinate pair options for Case 3:
(3, 4)
(3, -4)
(-3, 4)
(-3, -4)
(4, 3)
(4, -3)
(-4, 3)
(-4, -3)
Total ways = 8

Adding together the 3 cases, we get:
2 + 2 + 8 = 12.

The correct answer is E.
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by fskilnik@GMATH » Thu Nov 29, 2018 6:46 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

The equation of a circle in the x-y coordinate plane is x^2+y^2=25. How many points of the form (a,b), where a and b are integers, lie on this circle?

A. 4
B. 6
C. 8
D. 10
E. 12
$$?\,\,\,:\,\,\,\,\# \,\,\left( {x,y} \right)\,\,\,{\rm{integer}}\,\,{\rm{coordinates}}\,\,{\rm{solutions}}\,\,{\rm{for}}\,\,\,{x^2} + {y^2} = 25$$
$$\left| x \right| = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {x,y} \right) = \left( {0,5} \right)\,\,\,{\rm{or}}\,\,\,\left( {x,y} \right) = \left( {0, - 5} \right)$$
$$\left| x \right| = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{y^2} = 24 \ne \,\,{\rm{perfect}}\,\,{\rm{square}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{no}}\,\,{\rm{solutions}}\,$$
$$\left| x \right| = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{y^2} = 21 \ne \,\,{\rm{perfect}}\,\,{\rm{square}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{no}}\,\,{\rm{solutions}}\,\,\,$$
$$\left| x \right| = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{y^2} = 16\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\left( {x,y} \right) = \left( {3,4} \right)\,\,\,{\rm{or}}\,\,\,\left( {x,y} \right) = \left( {3, - 4} \right) \hfill \cr
\,\,\,{\rm{or}} \hfill \cr
\,\,\left( {x,y} \right) = \left( { - 3,4} \right)\,\,\,{\rm{or}}\,\,\,\left( {x,y} \right) = \left( { - 3, - 4} \right) \hfill \cr} \right.$$
$$\left| x \right| = 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{y^2} = 9\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\left( {x,y} \right) = \left( {4,3} \right)\,\,\,{\rm{or}}\,\,\,\left( {x,y} \right) = \left( {4, - 3} \right) \hfill \cr
\,\,\,{\rm{or}} \hfill \cr
\,\,\left( {x,y} \right) = \left( { - 4,3} \right)\,\,\,{\rm{or}}\,\,\,\left( {x,y} \right) = \left( { - 4, - 3} \right) \hfill \cr} \right.\,\,$$
$$\left| x \right| = 5\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{y^2} = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {x,y} \right) = \left( {5,0} \right)\,\,\,{\rm{or}}\,\,\,\left( {x,y} \right) = \left( { - 5,0} \right)$$

$$? = 12$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
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by Max@Math Revolution » Sun Dec 02, 2018 5:41 pm
=>

We need to find all pairs of integers (a,b) such that a^2 + b^2 = 25. These are:
(5,0), (4,3), (3,4), (0,5), (-3,4), (-4,3), (-5,0), (-4,-3), (-3,-4), (0,-5), (3,-4) and (4,-3).
Thus, there are 12 such points that lie on the circle.

Therefore, the answer is E.
Answer: E
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