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15 Nov 2018, 03:27
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
69% (01:31) correct
31%
(01:44)
wrong
based on 146
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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)
A (–1, 3√11)
B (0, –10)
C (–5, –7)
D (8, 6)
E (2, –4√6)
24 Nov 2018, 12:49
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parijit
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. [ square root of (-5)^2 + (-7)^2 = square root of 74]
24 Nov 2018, 18:56
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parijit
Attachment:
2018.11.24 circPythag.jpg [ 30.81 KiB | Viewed 4112 times ]
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), \(r=\sqrt{74}\).
\(\sqrt{74}\neq10\neq{r}\)
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 \(r=\sqrt{74}\neq{10}\)
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): \((2^2+(4√6)^2)=(4+(4*4*√6*√6)=(4+96)=100\)
