Question

i don't understand how to make it into a graphing form? how do you find the center and radius? (I'm confused because it has a 4x^2 and 4y^2 idk what to do with them ​

Answer 1

Answer:
`\bold{(x-2)^2+(y-3)^2=(1)/(4)}`

Center = (2, 3) radius =
`\bold{(1)/(2)}`

Explanation:

When both the x² and y² values are equal and positive, the shape is a circle. Complete the square to put the equation in format:

(x-h)² + (y-k)² = r² where

• (h, k) is the vertex
• r is the radius

1) Group the x's and y's together and move the number to the right side

4x² - 16x + 4y² - 24y = -51

2) Factor out the 4 from the x² and y²

4(x² - 4x ) + 4(y² - 6y ) = -51

3) Complete the square (divide the x and y value by 2 and square it)

`4[x^2-4x+\bigg((-4)/(2)\bigg)^2]+4[y^2-6y+\bigg((-6)/(2)\bigg)^2]=-51+4\bigg((-4)/(2)\bigg)^2+4\bigg((-6)/(2)\bigg)^2`

= 4(x - 2)² + 4(y - 3)² = -51 + 4(-2)² + 4(-3)²

= 4(x - 2)² + 4(y - 3)² = -51 + 4(4) + 4(9)

= 4(x - 2)² + 4(y - 3)² = -51 + 16 + 36

= 4(x - 2)² + 4(y - 3)² = 1

4) Divide both sides by 4

`(x-2)^2+(y-3)^2=(1)/(4)`

• (h, k) = (2, 3)
• 
  `r=(1)/(2)`