Question

The equation of a given circle in general form is x2+ y2 − 8x + 12y + 27 = 0. Write the equation in standard form, (x − h)2 + (y - k)2 = r2, by completing the squares in the equation. Show your work in a table.

Answer 1

The equation of the circle is
`x^(2) + y^(2) +8x+12y+27=0`

we take x-es and y-s 'close' to each other, and complete the square:


`x^(2)+8x + y^(2) +12y+27=0`

write the coefficients of the linear terms (the x and y with degree 1) as 2*'something', to see how to complete the square:


`x^(2)+2*4*x + y^(2) +2*6*y+27=0`

which means that
`x^(2)+2*4*x` needs
`4^(2)`

and
`y^(2) +2*6*y` needs
`6^(2)` to become perfect square trinomials:


`(x^(2)+2*4*x + 4^(2))-4^(2) + (y^(2) +2*6*y+ 6^(2))-6^(2) +27=0`


`(x+4)^(2)+ (y+6)^(2)=16+36-27`

`(x+4)^(2)+ (y+6)^(2)=25`


`(x-(-4))^(2)+ (y-(-6))^(2)= 5^(2)`


Answer:
`(x-(-4))^(2)+ (y-(-6))^(2)= 5^(2)`

Answer 2

Step Reason

x2 + y2 - 8x + 12y + 27 = 0 given

x2 - 8x + y2 + 12y = -27 Isolate the constant term

x2 - 8x + 16 + y2 + 12y + 36 = -27 + 16 + 36 Complete the square by adding . 16 and 36 to both sides.

(x2 - 2∙4∙x + 42) + y2 + 12y + 36 = 25 Group and rearrange terms in x.

(x - 4)2 + y2 + 12y + 36 = 25 (a - b)2 = a2 - 2ab + b2

(x - 4)2 + (y2 + 2∙6∙y + 62) = 25 Group and rearrange terms in y.

(x - 4)2 + (y + 6)2 = 25 (a + b)2 = a2 + 2ab + b2

(x - 4)2 + (y + 6)2 = 52 Take the square root to find r = 5.

Explanation:

this is the exact answer from plato

hope this helps a little bit :))