A circle has the equation x2+y2-8x+14y+40=0. Express this equation in center-radius form.

Question

Homework problem I can't figure out, there's two of these problems they're very similar but if someone can explain it to me step by step I think I can figure out the second one on my own.

 

X²+Y²-8X+14y+40=0

 

Thank you!

Answer 1

 

(x-h)²+(y-k)²=r² is the equation of a circle whose center is (h,k) and whose radius is r

x²+y²-8x+14y+40=0 is your equation which has to be put in the form above

(x²-8x+?)+(y²+14y+?)=-40

we need to find what the ?s are equal to

complete the square twice

take half of 8 and square it to get 16 for the first ?

take half of 14 and square it to get 49 for the second ?

replace the first ? with 16 and replace the second ? with 49 and don't forget to add 16 and 49 on the right side of the equation !!

(x²-8x+16)+(y²+14y+49)=-40+16+49

the two trinomials are perfect squares, rewrite them...

(x-4)²+(y-[-7])²=25 (you can write (y+7)² if you like; I wrote it so you would see the center as I explained above)

(x-4)²+(y-[-7])²=5² is the equation in center-radius form

(4,-7) is the center and 5 is the radius

 

 

 

Answer 2

Olesya -

Have you learned how to complete the square? 

 

First group like terms, and move the constant to the right hand side:

 

x² - 8x + ____ + y² + 14y + ____ = -40

 

in the first blank we need to put a number which would make the x's into a perfect square and 

in the second blank we need to put a number which would make the y's into a perfect square. 

 

to figure out what those would be, divide the x term by two: -8/2 = -4, and now square it: (-4)²=16 --> That's what we add to both sides of the equation. Now do this again for the y's 

 

14/2 = 7 ,  7²=49

 

now our equation looks like this: (x² - 8x + _16_) + (y² + 14y + _49_) = -40 +16 + 49

 

so what perfect squares did we make? (x² - 8x + 16) is (x - 4)²  

and (y² + 14y + 49) is (y +7)²

 

now our equation says (x - 4)² + (y + 7)² = 25

and we've achieved the center/radius form! 

 

Answer 3

Olesya,

 

The standard form of an equation of a circle can be represented as:

(x-h)² + (y-k)² = r²

 

      where the center is (h,k) and has a radius of r.

 

Let's rearrange your equation to get:

 

(x² - 8x) +(y² + 14y) + 40 = 0

 

Next, we need to get this in the form (x-h)² and (y-k)².  In order to do this let's complete the square for x and y variables by adding 16 to the x expression and adding 49 to the y expression.  To keep things balanced, we also need to subtract 16 and 49 from 40. 

 

(x² - 8x +16) + (y² + 14y + 49) + (40 - 16 - 49) = 0

 

Now, we can factor to get:

 

(x-4)² + (y+7)² - 25 = 0

 

Or:

 

(x-4)² + (y+7)² = 25

 

 

Hope this helps!