Problem with functions and relations?

Question

The relation given by the circle
x^2+y^2+6x-10y=15
does define two functions whose graphs are the upper and lower halves of the circle. Find expressions for these two functions.

Answer 1

First, put the circle's equation in to standard form:

x^2 + y^ +6x - 10y = 15

-->

[(x+3)^2 -3^2] + [(y-5)^2 - 5^2] = 15

-->

(x+3)^2 + (y-5)^2 = 15 +3^2 + 5^2

-->

(x+3)^2 + (y-5)^2 = 49

-->

(x - (-3))^2 + (y - (5))^2 = 7^2
. . . {this is standard form}

Now that we have the standard form we see that the circle is centered at (x,y) = (-3,5) and has a radius of 7.

The top half and bottom half will lie above/ below the line y = 5 (the line through the center).

So,the two funtcions are found by first transforming the standard form like this:

(x - (-3))^2 + (y - (5))^2 = 7^2
-->
(y - (5))^2 = 49- (x - (-3))^2

And then we can separate into two functions this way:

f1 --> y -5 = sqrt(49- (x - (-3))^2)
or, --> y = 5 +sqrt(49- (x - (-3))^2)

and

f2 --> y -5 = - (sqrt(49- (x - (-3))^2))
or, --> y = 5 -sqrt(49- (x - (-3))^2)

f1 will be the top half and f2 the bottom half.

Answer 2

You would have to solve for y to get two functions:

y^2-10y+25 = -x^2-6x+40

(complete the square)
x^2 + 6x + 9 + y^2 -10y + 25 -9 -25 = 15
(x+3)^2 + (y-5)^2 = 15 + 36
(y-5)^2 = 51-(x+3)^2
y-5 = +/- sqrt[ 51-(x+3)^2]
y = 5 +/- sqrt[ 51-(x+3)^2]

So these are the two functions.

Answer 3

y^2 -10y +25 = -x^2 -6x +40
(y-5)^2 = -x^2-6x+40
y-5 = +/- sqrt(-x^2-6x+40)
y = 5 + sqrt(-x^2-6x+40)
y = 5 - sqrt(-x^2-6x+40)

Answer 4

try x^2+6x=15+10y-y^2