Question

Which of the following could be points on the unit circle: A. (4/3, 4/5) B. (5/13, 12/13)

Answer 1

I think the answer is B 

Answer 2

Answer:

Option (2) and (4) are correct.

$(\frac{5}{13},\frac{12}{13} )$ and  $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ are points on the unit circle.

Step-by-step explanation:

Given : Some points of circles.

We have to choose which points could be points on the unit circle.

We know, the equation of circle is

$x^2+y^2=r^2$  ..........(1)

where r is radius of circle.

We check each point for x and y values on the equation of circle and see which point gives radius = 1

Thus,

1) $(\frac{4}{3},\frac{4}{5} )$

Put  in LHS of  (1) , we have,

$(\frac{4}{3})^2+(\frac{4}{5})^2$

Simplify, we have,

$=\frac{16}{9}+\frac{16}{25}$

$=\frac{544}{225}\neq 1$

Thus,  $(\frac{4}{3},\frac{4}{5} )$ is not a point on the unit circle.

2) $(\frac{5}{13},\frac{12}{13} )$

Put  in LHS of  (1) , we have,

$(\frac{5}{13})^2+(\frac{12}{13})^2$

Simplify, we have,

$=\frac{25}{169}+\frac{144}{169}$

$=\frac{169}{169}= 1$

Thus, $(\frac{5}{13},\frac{12}{13} )$ is a point on the unit circle.

3) $(\frac{1}{3},\frac{2}{3} )$

Put  in LHS of  (1) , we have,

$(\frac{1}{3})^2+(\frac{2}{3})^2$

Simplify, we have,

$=\frac{1}{9}+\frac{4}{9}$

$=\frac{5}{9}\neq 1$

Thus,  $(\frac{1}{3},\frac{2}{3} )$ is not a point on the unit circle.

4)  $(\frac{6}{7},\frac{\sqrt{13}}{7} )$

Put  in LHS of  (1) , we have,

$(\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2$

Simplify, we have,

$=\frac{36}{49}+\frac{13}{49}$

$=\frac{49}{49}= 1$

Thus,  $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ is a point on the unit circle.

Thus, $(\frac{5}{13},\frac{12}{13} )$ and  $(\frac{6}{7},\frac{\sqrt{13}}{7} )$ are points on the unit circle.