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.

ANSWER CExplanationFor 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 CThe 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):

\((2^2+(4√6)^2)=(4+(4*4*√6*√6)=(4+96)=100\)