Answer to Question #266065 in Algebra for Paul Carpenter

Question

Discussion Assignment

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

This determines the relation between x and y as x and y are chosen in such a way that these form a circle of unit radius.

Hence we can say that x and y are related to each other i.e.

$(x-a)^2 + (y-b)^2 = 1$ where (a,b) is the center of the circle.

If (a,b) is (0,2) then $x^2 + (y-2)^2 = 1$

then $x=g(y) = \sqrt{1-(y-2)^2}$ this is relation which shows the dependency of x on y

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

The domain for g(y) is [1,3] as y will take those values for which x is real. and that for h(x) is [0,1]

Point(0,2) is related by R as it will represent a circle of unit radius.