Question

Consider the equation x^2+(y-2)^2=1 and the relation "(x, y) r (0, 2)", where r is read as "has distance 1 of". For example, "(0, 3) r (0, 2)", that is, "(0, 3) has distance 1 of (0, 2)". This relation can also be read as "the point (x, y) is on the circle of radius 1 with center (0, 2)". In other words: "(x, y) satisfies this equation x^2+(y-2)^2=1, if and only if, (x, y) r (0, 2)". Does this equation determine a relation between x and y? can the variable x can be seen as a function of y, like x=g(y)? can the variable y be expressed as a function of x, like y= h(x)? if these are possible, then what will be the domains for these two functions? what are the graphs of these two functions? are there points of the coordinate axes that relate to (0, 2) by means of r?.

Answer 1

• x can be written as a function of y but the reverse is not the case.
• The domain of the function g(y) occurs at y ≤ 3.
• Also, since the center is at (0,2) hence the points of the coordinate axes that relate to (0, 2) by means of r

Equation of a circle

Given the equation of a circle expressed as $x^2+(y-2)^2=1$

The standard form of the equation of a circle is expressed as $(x-a)^2+(y-b)^2=r^2$

where:

• (a, b) is the center of the circle
• r is the radius of the circle

Compared with the given equation, you can see that the center of the circle is (0, 2), and the radius s 1 unit

Write the variable x as a function of "y" as shown:

$x^2+(y-2)^2=1\\x^2 = 1 - (y-2)^2\\x = \sqrt{1 - (y-2)^2}$

This shows that x can be written as a function of y but the reverse is not the case.

The possible domain of g(y) occurs when the function (y-2)² ≤ 1

(y-2)² ≤ 1

(y-2)² ≤ 1

(y-2)²≤ 1

y - 2≤ 1

y ≤ 3

Hence the domain of the function occurs at y ≤ 3.

Also, since the center is at (0,2) hence the points of the coordinate axes that relate to (0, 2) by means of r

Find the graph of g(y) attached below:

Learn more on equation of a circle here: https://brainly.com/question/1506955