All of the following xy-coordinate points lie on the circumference of

Question

All of the following xy-coordinate points lie on the circumference of a circle whose radius is 10 and whose center is the (x,y) point (0,0) EXCEPT:

A (–1, 3√11)
B (0, –10)
C (–5, –7)
D (8, 6)
E (2, –4√6)

Ohbebby · Answer

Just use the distance formula and you are done!

distance formula = Square root of (x2-x1)^2 + (y2-y1)^2 where the 2 points are in the form of x1,y1 and x2,y2

So by the formula All the options on the circumference should give the value = 10.

Only C does not give the value 10.

generis · Answer

2018.11.24 circPythag.jpg We can also use the equation of a circle , which is derived from the Pythagorean theorem. All (x,y) points on the circle will satisfy the equation. • The general equation of a circle with center (0,0) is x^2 + y^2 = r^2 r=10 x^2 + y^2 = 10^2 x^2 + y^2 = 100 • Each point (x,y) must satisfy the equation (1) Options B (y axis intercept) and D (a 3-4-5 triangle) obviously do satisfy the equation and thus lie on the circle. B (0,10): ( 0^2 + (-10)^2)=(0+100)=100 D (8,6): (8^2+6^2)=(64+36)=100 (2) A, C, and E remain. Test (C). Its numbers are round and not part of a Pythagorean triplet C (5,7): (5^2+7^2)=(25+49)=74 r^2 = 74 r^2 should = 100. In (C), r=\sqrt{74} . \sqrt{74} eq10 eq{r} Point C does NOT lie on the circle. ANSWER C Explanation For every point (x,y) that lies on the circle, if we draw a perpendicular line from x to the axis, we form a right triangle. See diagram The radius of the circle is the hypotenuse of a right triangle with base x and height y . We we use the Pythagorean theorem a^2 + b^2 = c^2 ; the names of variables are different. a = base of right triangle = x b = height of right triangle = y c = the hypotenuse of the right triangle = r In this case, radius = 10 , so r^2 = 100 Look at the answer choices. Which pair, squared, does NOT sum to 100? • B and D are eliminated immediately. B (0, –10) is where the circle intersects the y-axis Or plug the coordinates into the equation. x^2 + y^2 = r^2 0^2+(-10)^2 = 10^2 0 + 100 = 100 D (8, 6) is a 3-4-5 => (6-8-10) triangle. The radius is 10. That works. OR x = 8, y = 6, r = 10 8^2 + 6^2 = 10^2 64 + 36 = 100 Options A, C, and E remain. • Before dealing with radicals in (A) and (E) . . . C (5,7) looks very suspicious. 5 and 7 are not part of a Pythagorean triplet, let alone one whose radius is 10 Test the coordinates; x and y must satisfy the equation If they do not, then Point C is not on the circle 5^2+7^2=r^2 (25+49)=74=r^2 r^2 = 74 Stop. r^2 must = 100 It does not. This point does not lie on the circle. Or r=\sqrt{74} eq{10} Point C does not lie on the circle ANSWER C The coordinates of points A and E, squared, will sum to 100, such that r^2=100 and r=10 Checking A, for example A (–1, 3√11): x^2+y^2=r^2 (-1)^2+(3√11)^2 = (1 + (3*3*√11*√11)) = (1 + (9*11)) = (1+99) = 100 That works. Point A lies on the circle. Point E also lies on the circle. E (2, 4√6): (2^2+(4√6)^2)=(4+(4*4*√6*√6)=(4+96)=100

ShreyaKohli · Answer

If the circle is centred at (0,0), then by the formula :

(x-a)^2 + (y-b)^2 = r^2 , where (a,b) is the origin and (x,y) are the points that lie on the circumference 
x^2+y^2=r^2 (Since a,b = 0,0)

So look for the points from the answer choices which when squared equal to 10^2, i.e. 100.
C is the only option which does not suffice this requirement as (-5)^2+(-7)^2 = 84

All of the following xy-coordinate points lie on the circumference of

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

GMAT Problem Solving (PS) Questions

All of the following xy-coordinate points lie on the circumference of

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia