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Write the equation of the circle in standard form. Identify the center point and the radius, and then sketch it's graph for:
x²+y²+4x-6y-3=0

Sagot :

Answer:

The center of the circle is at (-2, 3) and has a radius of [tex]\sqrt{13}[/tex].

Step-by-step explanation:

Circles

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]

Completing the Square

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]

Solving the Problem

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]

Graphing

Start by making a dot at the center of the circle, then draw four dots that are about 3.6 units

  • above
  • below
  • left
  • right

of the center.

Lastly, draw curved lines that connect those four dot together, forming the circle.

The image below is used by Geogebra.

View image zarahaider4211
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