Answer:
Standard form: [tex](x-\frac{1}{2})^2 + (y-0)^2 = 15[/tex]
Center: [tex](\frac{1}{2}, 0)[/tex]
Radius: [tex]r =\sqrt{15}[/tex]
Step-by-step explanation:
The equation of a circle in the standard form is
[tex](x-h)^{2} + (y-k)^{2} = r^{2}[/tex]
Where the point (h, k) is the center of the circle
To transform this equation [tex]4x^{2} -4x + 4y^{2} - 59 = 0[/tex] this equation in the standard form we use the method of square.
First, we group similar variables
[tex](4x^{2} -4x) + (4y^{2}) - 59 = 0[/tex]
Divide both sides of equality by 4
[tex](x^{2} -x) + (y^{2}) - 14.75 = 0[/tex]
Now we complete square for variable x.
Take the coefficient "b" that accompanies the variable x and divide by 2. Then, elevate the result to the square:
[tex]b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2= (-\frac{1}{2})^2 = \frac{1}{4}[/tex]
Now add [tex](\frac{b}{2})^2[/tex] on both sides of the equality
[tex](x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})[/tex]
Factor the expression and simplify the independent terms
[tex](x-\frac{1}{2})^2 + (y^{2}) = 15[/tex]
[tex](x-\frac{1}{2})^2 + (y-0)^2 = 15[/tex]
Then
[tex]h =\frac{1}{2}\\\\k=0[/tex]
and the center is [tex](\frac{1}{2}, 0)[/tex]
radius [tex]r =\sqrt{15}[/tex]
