Question

Using the method of completing the square, put each circle into the form

Answer 1

Answer:

Standard form: $(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Center: $(\frac{1}{2}, 0)$

Radius: $r =\sqrt{15}$

Step-by-step explanation:

The equation of a circle in the standard form is

$(x-h)^{2} + (y-k)^{2} = r^{2}$

Where the point (h, k) is the center of the circle

To transform this equation $4x^{2} -4x + 4y^{2} - 59 = 0$ this equation  in the standard form we use the method of square.

First, we group similar variables

$(4x^{2} -4x) + (4y^{2}) - 59 = 0$

Divide both sides of equality by 4

$(x^{2} -x) + (y^{2}) - 14.75 = 0$

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:

$b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2= (-\frac{1}{2})^2 = \frac{1}{4}$

Now add $(\frac{b}{2})^2$ on both sides of the equality

$(x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})$

Factor the expression and simplify the independent terms

$(x-\frac{1}{2})^2 + (y^{2}) = 15$

$(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Then

$h =\frac{1}{2}\\\\k=0$

and the center is $(\frac{1}{2}, 0)$

radius $r =\sqrt{15}$