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: $\bold{(x-2)^2+(y-3)^2=\dfrac{1}{4}}$

               Center = (2, 3)          radius = $\bold{\dfrac{1}{2}}$

Step-by-step 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)

    $4[x^2-4x+\bigg(\dfrac{-4}{2}\bigg)^2]+4[y^2-6y+\bigg(\dfrac{-6}{2}\bigg)^2]=-51+4\bigg(\dfrac{-4}{2}\bigg)^2+4\bigg(\dfrac{-6}{2}\bigg)^2$

  = 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

   $(x-2)^2+(y-3)^2=\dfrac{1}{4}$

• (h, k) = (2, 3)
• $r=\dfrac{1}{2}$