Answer:
The center of the circle is at (-2, 3) and has a radius of [tex]\sqrt{13}[/tex].
Step-by-step explanation:
The standard equation for a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex],
where (h, k) is the center of the circle and r is its radius.
[tex]\dotfill[/tex]
When a square is completed, the quadratic can be factored where the factors are the same and can thus be written as a single factored squared.
For example,
[tex]x^2+10x+25=(x+5)(x+5)=(x+5)^2[/tex].
All completed squares should end as [tex](x-s)^2[/tex] for any real number s. This should look familiar to the circle equation.
To complete the square, we must add a new "c" value. To find that "c" value, take the squared value of one-half of b.
Make sure that when you add it into the quadratic it's also added to the other side of the equation.
I.e.
[tex]ax^2+bx+d+\left(\dfrac{b}{2} \right)^2= \left(\dfrac{b}{2} \right)^2[/tex],
where d doesn't make a complete square.
All that there's left is to factor between the x squared term, the bx, and the new c value we added.
[tex]\hrulefill[/tex]
We need to complete the square twice, for the x and y terms, for our initial equation to have the standard circle form.
We can start by organizing the terms.
[tex]x^2+4x+y^2-6y-3=0[/tex]
We can even bring the 3 onto the other side to make things easier.
[tex]x^2+4x+y^2-6y=3[/tex]
We use the general formula to find the new c value for the x and y terms:
[tex]c_x=\left(\dfrac{4}{2} \right)^2=(2)^2=4[/tex],
[tex]c_y=\left(\dfrac{-6}{2} \right)^2=(-3)^2=9[/tex].
We add the 4 and 9 to both sides of the equation so that they the equation still makes logical sense.
[tex]x^2+4x+4+y^2-6y+9=3+4+9[/tex]
Now we factor!
[tex](x+2)(x+2)+(y-3)(y-3)=13[/tex]
[tex](x+2)^2+(y-3)^2=13[/tex]
So, our circle has a center as (-2, 3) and a radius of [tex]\sqrt{13}[/tex]!
[tex]\dotfill[/tex]
Start by making a dot at the center of the circle, then draw four dots that are about 3.6 units
of the center.
Lastly, draw curved lines that connect those four dot together, forming the circle.
The image below is used by Geogebra.
