Graphing a Relation: x^2 + y^2 - 10x + 4y + 13 = 0

[timpickens]

I can't figure out how to put this in to y= form.

x^2 + y^2 - 10x + 4y + 13 = 0

It also asks to indentify the graph and give the radius, center, vertices, and foci.

I've tried several different methods for solving this but haven't come out with any working results.

Any help would be immensely appreciated!

 

[stapel]

I can't figure out how to put this in to y= form.
x^2 + y^2 - 10x + 4y + 13 = 0

I'm surprised the exercise required that of you, because the equation of a circle isn't a function. But since those were the instructions...

. . . . .x² + y² - 10x + 4y + 13 = 0

. . . . .y² + 4y + (x² - 10x + 13) = 0

Then apply the Quadratic Formula, with a = 1, b = 4, and c = x² - 10x + 13.

It also asks to indentify the graph and give the radius, center, vertices, and foci.

This part sounds much more standard. Since the equation is that of a circle, follow the procedure they showed you for completing the square. Then read off the values of the center and radius, and draw the circle.

Eliz.

 

[galactus]

Try grouping and completing the square.

\(\displaystyle \L\\x^{2}+y^{2}-10x+4y=-13\)

\(\displaystyle \L\\(x^{2}-10x+?)+(y^{2}+4y+?)=-13+?+?\)

After you complete the squares, factor each. See?

 

[tkhunny]

I can't figure out how to put this in to y= form.

That, really, is a very odd question. Were you asked to do that? Since this probably isn't a "function", it seems unlikely that there is such a form. There may be such a form if you count the two pieces as only one.

 

[timpickens]

Ah... I misunderstood the problem, sorry, I didn't actually need to put it in to y= form. But I was still stumped on the rest as my math book didn't seem to explain the concept in much detail. Thanks so much, guys!