Dint get this

[email protected] wrote:A circle with a diameter 10 is centered on the origin, which of the following points are on the circle
a)(5,0)
b)(5,5)
c)(-3,4)
d) (1,-2sqrt6)
e) (-2, sqrt6)

I think you mean "sqrt" (not "underroot")
I have edited the question accordingly.

NOTE: This question is un-GMAT-like because it allows for more than 1 correct answer. That said, it could possibly appear in the IR section.

Okay, the first thing to recognize is that the circle has radius 5.

If a circle of radius is centered on the origin, any point (x,y) on the circle must be such that x^2 + y^2 = 5^2. Here's why . . .

Let's take a point on the circle, say (-3, 4), notice that we can create a right triangle.

Since we have a right triangle, Pythagoras tells us that (-3)^2 + 4^2 = 5^2
In general, any point (x,y) on the circle must be such that x^2 + y^2 = 5^2

Now we'll check the answer choices.
a)(5,0): does 5^2 + 0^2 = 5^2? YES! So, this point is on the circle.
b)(5,5): does 5^2 + 5^2 = 5^2? NO So, this point is not on the circle.
c)(-3,4): does (-3)^2 + 4^2 = 5^2? YES! So, this point is on the circle.
d) (1,-2sqrt6): does 1^2 + (-2sqrt6)^2 = 5^2? YES! So, this point is on the circle.
e) (-2, sqrt6): does (-2)^2 + (sqrt6)^2 = 5^2? NO So, this point is not on the circle.

Answers: [spoiler]A, C & D[/spoiler]

Cheers,
Brent

v1nc3nz0 wrote:

Brent@GMATPrepNow wrote:

[email protected] wrote:A circle with a diameter 10 is centered on the origin, which of the following points are on the circle
a)(5,0)
b)(5,5)
c)(-3,4)
d) (1,-2sqrt6)
e) (-2, sqrt6)

I think you mean "sqrt" (not "underroot")
I have edited the question accordingly.

NOTE: This question is un-GMAT-like because it allows for more than 1 correct answer. That said, it could possibly appear in the IR section.

Okay, the first thing to recognize is that the circle has radius 5.

If a circle of radius is centered on the origin, any point (x,y) on the circle must be such that x^2 + y^2 = 5^2. Here's why . . .

Let's take a point on the circle, say (-3, 4), notice that we can create a right triangle.

Since we have a right triangle, Pythagoras tells us that (-3)^2 + 4^2 = 5^2
In general, any point (x,y) on the circle must be such that x^2 + y^2 = 5^2

Now we'll check the answer choices.
a)(5,0): does 5^2 + 0^2 = 5^2? YES! So, this point is on the circle.
b)(5,5): does 5^2 + 5^2 = 5^2? NO So, this point is not on the circle.
c)(-3,4): does (-3)^2 + 4^2 = 5^2? YES! So, this point is on the circle.
d) (1,-2sqrt6): does 1^2 + (-2sqrt6)^2 = 5^2? NO So, this point is not on the circle.
e) (-2, sqrt6): does (-2)^2 + (sqrt6)^2 = 5^2? NO So, this point is not on the circle.

Answers: [spoiler]A & C[/spoiler]

Cheers,
Brent

i think that also D is correct:
(1,-2sqrt6): does 1^2 + (-2sqrt6)^2 = 1^2 + (-2)*(-2)*(sqrt6)*(sqrt6)= 1 + 4 * 6 = 25 = 5^2

Am i wrong ?

Good catch!
You're absolutely right - I've edited my response accordingly.

Thanks and cheers,
Brent