anonymous
  • anonymous
An equation of a circle is x^2 + y^2 + 10x – 6y + 18 = 0. Show all your work in determining the center and radius of this circle. In complete sentences, explain the procedure used.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
abb0t
  • abb0t
The form of the circle is: \[(x-h)+(y-k) = r^2 \] where (h,k) is the center of the circle, and r is the radius. What you want to do is rearrange it to get it in that form. Use algebra to group (not combine) like terms. Then complete the square
anonymous
  • anonymous
what is h and k in this equation?
abb0t
  • abb0t
Well, that's up to you to figure out. First, group like terms: \[(x^2+10x)+(y^2-6y)+18 = 0\] Then, proceed to complete the square for x and y to get it in the form I stated earlier. Does that make more sense?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

abb0t
  • abb0t
To complete the square: \[ax^2+bx + e\] you add to both sides (including product side): \[(\frac{ b }{ 2 })^2 \]
anonymous
  • anonymous
i know how to complete the square
abb0t
  • abb0t
gotcha.

Looking for something else?

Not the answer you are looking for? Search for more explanations.