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01 Jan 2008, 03:11
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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2. In the xy plane, point (r,s) lies on the circle with centre at the origin. What is the value of r^2 + s^2 ?
1. The circle has radius 2.
2. The point ( v2,-v2) lies on the circle.
1. The circle has radius 2.
2. The point ( v2,-v2) lies on the circle.
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01 Jan 2008, 09:33
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OK, so we have a circle with it's center at the origin. Point (r,s) is on the circle and we want to know the value of r^2 + s^2.
I just found this on google but it happens to work perfectly for us! Here is a circle centered around the origin. Now imagine a point somewhere on that circle. From that point extend a line to the X-axis OR extend a line to the Y-axis. Now extend a line to the center of the circle. What we have done is create a right triangle with one side the length of R and the other side the length of S. The line extending to the center of the circle is the hypotenuse as well as the radius of the circle! Using the Pythagorean theorem we know that A^2 + B^2 = C^2 in this sort of triangle. Using this triangle R and S are A and B, so r^2 + s^2 is simply the square of the radius. For any point on this circle r^2 + s^2 must equal to the square of the radius. Now with this information in hand we can read the statements knowing that all we need to find is the radius.
1. The circle has radius 2.
They came right out and told us! SUFFICIENT
2. The point ( v2,-v2) lies on the circle. (I assume these are sqrt(2), -sqrt(2))
These coordinates are the two legs of our triangle So square both of them and add them together...you get 4. The sqrt(4) = 2 which is our radius, but you don't even need to find that. Just knowing that the coordinates squared add up to 4 tell us that they'll do the same everywhere on the circle.
SUFFICIENT
Answer D
PS and if the point were to fall on the X or Y axis we can't draw a triangle but we know the coordinates are 0 and 2. or 0 and -2. Either way they add up to 4 like everywhere else.
I just found this on google but it happens to work perfectly for us! Here is a circle centered around the origin. Now imagine a point somewhere on that circle. From that point extend a line to the X-axis OR extend a line to the Y-axis. Now extend a line to the center of the circle. What we have done is create a right triangle with one side the length of R and the other side the length of S. The line extending to the center of the circle is the hypotenuse as well as the radius of the circle! Using the Pythagorean theorem we know that A^2 + B^2 = C^2 in this sort of triangle. Using this triangle R and S are A and B, so r^2 + s^2 is simply the square of the radius. For any point on this circle r^2 + s^2 must equal to the square of the radius. Now with this information in hand we can read the statements knowing that all we need to find is the radius.
1. The circle has radius 2.
They came right out and told us! SUFFICIENT
2. The point ( v2,-v2) lies on the circle. (I assume these are sqrt(2), -sqrt(2))
These coordinates are the two legs of our triangle So square both of them and add them together...you get 4. The sqrt(4) = 2 which is our radius, but you don't even need to find that. Just knowing that the coordinates squared add up to 4 tell us that they'll do the same everywhere on the circle.
SUFFICIENT
Answer D
PS and if the point were to fall on the X or Y axis we can't draw a triangle but we know the coordinates are 0 and 2. or 0 and -2. Either way they add up to 4 like everywhere else.
01 Jan 2008, 19:13
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I figured out that the 1st statement is sufficient but I had a problrm with understanding the point (v2,-v2)...you assumed this as (sqrt2, -sqrt2) still didn't fathom that but i guess that I can find this sufficient by assuming the position of the point on X or on Y axis. thx.
OA is D.
OA is D.
