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

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Joined: 02 Sep 2018
Posts: 65
All of the following xy-coordinate points lie on the circumference of  [#permalink]

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15 Nov 2018, 02:27
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35% (medium)

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66% (01:04) correct 34% (01:05) wrong based on 56 sessions

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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)
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Joined: 21 Jun 2018
Posts: 37
Re: All of the following xy-coordinate points lie on the circumference of  [#permalink]

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24 Nov 2018, 11:49
parijit wrote:
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)

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]
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All of the following xy-coordinate points lie on the circumference of  [#permalink]

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24 Nov 2018, 17:56
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parijit wrote:
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)

Attachment:

2018.11.24 circPythag.jpg [ 30.81 KiB | Viewed 188 times ]

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), $$r=\sqrt{74}$$.
$$\sqrt{74}\neq10\neq{r}$$

Point C does NOT lie on the circle.

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

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): $$(2^2+(4√6)^2)=(4+(4*4*√6*√6)=(4+96)=100$$
All of the following xy-coordinate points lie on the circumference of &nbs [#permalink] 24 Nov 2018, 17:56
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