show that the equation 4x^2+4y^2-4x-12y+1=0 represents a circle. find its center and radius
You need to complete to square for your x's and your y's to put it into standard form for a circle.
get in in the form: (x-a)^2 + (y-b)^2 = r^2 where r is the radius, and (a,b) the center
First, divide each term by 4
\[x^2-x+y^2-3y=-\frac{1}{4}\]
Now add 1/4 and 9/4 to the x an y parts respectively and to the right side
k i divided each term by 4 and i got \[x^2+y^2+x-3y+1/4=0\] am i ryt?
\[x^2+x+\frac{1}{4}+y^2-3y+\frac{9}{4}=-\frac{1}{4}+\frac{1}{4}+\frac{9}{4}\]
Now factor the two trinomial squares and add the like terms to determine the center and radius.
k could u pls explain how u got 9/4?
The y coefficient is 3. (1/2)(3)=3/2 and (3/2)^2=9/4
k what do i do next to get center nd radius?
Now factor the two trinomial squares and add the like terms to determine the center and radius.
k i got x^2+y^2+x-3y+5/2=-5/2 am i ryt?
mertsj how do i find center and radius from here
\[(x+\frac{1}{2})^2+(y-\frac{3}{2})^2=\frac{9}{4}\]
Center is (-1/2, 3/2) Radius is 3/2
how did get it
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