Coordinate geometry, circles

[sapuna]

Just to let you know , I guessed this one when doing the tests and afterwards correctly solved it when I had the time. However , i did so by using a compass and some coordinate paper. On test day I won`t have the time. Obviously there is some circles property I don`t know , so tell it to me if possible

Question : What are the coordinates of point A

1) A is 2 units away from point (3;4)
2) A is 3 units away from point (0;0)

Obviously neither statement alone is sufficient because they give circles with too many points. Using statements 1 and 2 ( and a compass nad paper ) I indeed found out that the two circles intersect at only 1 point therefore statements 1 and 2 combined are enough.

However, when you have to solve it for 2 minutes how do you do it ? I tried finding the distance between the centers of the two circles : square root of (3-0)^2 + (4-0)^2 which is square root of 16 + 9 = 5. Two circles can have no common point, 1 common point or 2 common points. How do we know they have exactly 1 common point from the given info ?

[GMATGuruNY]

Question : What are the coordinates of point A

1) A is 2 units away from point (3;4)
2) A is 3 units away from point (0;0)

Obviously neither statement alone is sufficient because they give circles with too many points.

DRAW.
When the statements are combined, we get the following figure:
Figure 1:

R, S and T form the sides of a 3-4-5 triangle.
Thus, RS = 5.

Statement 1: Point A must be 3 units away from R, implying that RA=3.
Statement 2: Point A must be 2 units away from S, implying that AS=2.
Since RS=5, it is possible that RA=3 and AS=2 only if point A lies on side RS, as in the figure above.
Thus, the coordinates of point A can be determined.
SUFFICIENT.

The correct answer is C.

If point A does NOT lie on side RS, we get a figure like the following:


The sum of the lengths of any 2 sides of a triangle must be GREATER than the length of the 3rd side.
Thus, in ∆RAS, RA + AS > 5.
If RA + AS > 5,then it is not possible that RA=3 and AS=2.
Implication:
In order that RA=3 and AS=2, point A MUST lie on side RS of 3-4-5 ∆RST, as shown in Figure 1 above.

[Brent@GMATPrepNow]

What are the coordinates of point A?

1) A is 2 units away from point (3;4)
2) A is 3 units away from point (0;0)

IMPORTANT: For geometry Data Sufficiency questions, we are typically checking to see whether the statements "lock" a particular angle, length, or shape into having just one possible measurement. This concept is discussed in much greater detail in our free video: https://www.gmatprepnow.com/module/gmat- ... cy?id=1103

As you have suggested, the trick here is to determine whether the 2 circles intersect at one point or two points (as shown here).


ASIDE: For this question, it won't be the case that the 2 circles do not intersect. Otherwise, the two statements would contradict each other. That is, if the 2 circles do not intersect, point A CANNOT BE 2 units away from point (3,4) AND 3 units away from point (0,0).
I expound on this in our free video titled "Useful Contradictions": https://www.gmatprepnow.com/module/gmat- ... cy?id=1104

Okay, here's my solution:

Target question: What are the coordinates of point A?

Statement 1: A is 2 units away from point (3,4)
There are MANY MANY points that are 2 units away from (3,4). In fact, all points on a circle centered at (3,4) with radius 2 will satisfy this condition.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: A is 3 units away from point (0,0)
There are MANY MANY points that are 4 units away from (0,0). In fact, all points on a circle centered at (0,0) with radius 3 will satisfy this condition.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
First, (0,0) and (3,4) are 5 units apart.


This means that the two circles (with radii of 2 and 3) will intersect at ONLY ONE POINT.

So, we have LOCKED IN point A to having just one location.
Do we need to find the actual coordinates of A?
No. We need only recognize that there is only one possible point for A.
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent

[sapuna]

I`m not sure I got this one. The three sides of the triangle are with measurments 3,4,5 . How did we knwo from that that the two circles intersect at just 1 point ? When will they intersect at two points ?

[GMATinsight]

I`m not sure I got this one. The three sides of the triangle are with measurments 3,4,5 . How did we knwo from that that the two circles intersect at just 1 point ? When will they intersect at two points ?

3-4-5 is a Right angle Triangle
3^2 + 4^2 = 5^2

The Distance Between (0,0) and (3,4) is 5 Units

and If you draw circles with radius 2 and 3 from points (3,4) and (0,0) Respectively then the two points are bound to meet at just one point (As shown in the figure I have posted below) because 2+3=5

The two circles drawn will intersect at two points if the sum of radii of the two circles is greater than 5 (in this particular scenario)
i.e. x+y>5 where x and y are radii of the circles drawn from the end points of line 5 unit long.

[Brent@GMATPrepNow]

I`m not sure I got this one. The three sides of the triangle are with measurments 3,4,5 . How did we knwo from that that the two circles intersect at just 1 point ? When will they intersect at two points ?

The distance from (3,4) to point A is 2 (since the radius of that circle is 2) and the distance from (0,0) to point A is 3 (since the radius of that circle is 3)

Since the TOTAL distance from (0,0) to (3,4) is 5, we can see that point A is the ONE point on the line segment from (0,0) to (3,4)

Cheers,
Brent

[GMATinsight]

Check the image below for understanding One and two point of intersection cases

[GMATGuruNY]

I`m not sure I got this one.

In my post above, I've added a second figure to clarify the reasoning.

[sapuna]

Thank you both guys so much. I knew from the beginning the sides of the triangles; the part I didn`t understand was the circles. Now I know that if the sum of the radii is bigger than the distance between the two circles , there will be 2 points in which the circles intersect

[Matt@VeritasPrep]

When in geometric doubt, why not turn to algebra?

Suppose A is the coordinate (x,y).

S1 tells us that

2 = √((3-x)² + (4-y)²)

This is insufficient, as it has at least two answers.

S2 tells us that

3 = √(x² + y²)

which also has at least two answers.

Together, however, we have

9 = x² + y² (from the second equation)

and

4 = (3-x)² + (4-y)²
4 = 9 - 6x + x² + 16 - 8y + y²

Combining the equations gives us

4 = 25 - 6x - 8y + 9, or

6x + 8y = 30, or

3x + 4y = 15

Combining this equation with x² + y² = 9 gives us one unique solution: x = 9/5, y = 12/5.

[Brent@GMATPrepNow]

When in geometric doubt, why not turn to algebra?

Suppose A is the coordinate (x,y).

S1 tells us that

2 = √((3-x)² + (4-y)²)

This is insufficient, as it has at least two answers.

S2 tells us that

3 = √(x² + y²)

which also has at least two answers.

Together, however, we have

9 = x² + y² (from the second equation)

and

4 = (3-x)² + (4-y)²
4 = 9 - 6x + x² + 16 - 8y + y²

Combining the equations gives us

4 = 25 - 6x - 8y + 9, or

6x + 8y = 30, or

3x + 4y = 15

Combining this equation with x² + y² = 9 gives us one unique solution: x = 9/5, y = 12/5.

It might be worth noting (to those who may be beginning their prep and unaware of the GMAT syllabus) that the test-makers DO NOT expect you to know that 2 = √((3-x)² + (4-y)²) is the equation of a circle with radius 4 and center (3,4)

BTW Matt, how do we know that the system (3x + 4y = 15 and x² + y² = 9) has only one solution?

Cheers,
Brent

[Matt@VeritasPrep]

It might be worth noting (to those who may be beginning their prep and unaware of the GMAT syllabus) that the test-makers DO NOT expect you to know that 2 = √((3-x)² + (4-y)²) is the equation of a circle with radius 4 and center (3,4)

BTW Matt, how do we know that the system (3x + 4y = 15 and x² + y² = 9) has only one solution?

Brent, you're always keeping me on my toes!

I meant that as a general, GMAT-friendly distance equation, not a fancypants "the circle with radius r and center (h,k) has the equation BLAH" equation. The distance from (x,y) to (3,4) is √((3-x)² + (4-y)²), and since that distance is 2, we have 2 = √((3-x)² + (4-y)²).

As for the single solution, how about this? Assume 3x + 4y = 15 intersects the circle. This line will be tangent to the line y = (4/3)x at a point (x,y) that satisfies both equations. Solving y = (4/3)x and 3x + 4y = 15 gives us y = 12/5, x = 9/5. But these values satisfy x² + y² = 9, and y = (4/3)x goes through the center of the circle x² + y² = 9, so 3x + 4y = 15 is tangent to the circle and we only have one set of solutions.

[KristenH88]

What did they mean by "units" away? It's not spaces; mathematically I just don't understand the terminology. Breaking down what's being asked is my largest hurdle.

[Brent@GMATPrepNow]

Ah, good call about the distance formula vs the circle equation.

I think you system-solving solution should come with a "do not try this at home" warning

Cheers,
Brent

[Brent@GMATPrepNow]

What did they mean by "units" away? It's not spaces; mathematically I just don't understand the terminology. Breaking down what's being asked is my largest hurdle.

In the coordinate plane, "units" can be thought of as spaces. So, for example, the distance between (0,0) and (0,1) is 1 unit.
Likewise, (3,0) is 5 units away from (8,0)

Does that help?

Cheers,
Brent