Respuesta :
Answer:
Part A) [tex](x-1)^{2}+(y-2)^{2}=3^{2}[/tex]
Part B) The coordinates of the center are [tex](h,k)=(1,2)[/tex] and the radius is equal to [tex]r=3\ units[/tex]
Part C) The graph in the attached figure
Step-by-step explanation:
we have
[tex]x^{2} +y^{2}-2x-4y-4=0[/tex]
Part A) Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}-2x) +(y^{2}-4y)=4[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side
[tex](x^{2}-2x+1) +(y^{2}-4y+4)=4+1+4[/tex]
[tex](x^{2}-2x+1) +(y^{2}-4y+4)=9[/tex]
Rewrite as perfect squares
[tex](x-1)^{2}+(y-2)^{2}=3^{2}[/tex] ------> Is a circle
Part B) If this is a circle, state the coordinates of the center and give the radius.
we know that
The equation of the circle in standard form is equal to
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
where
(h,k) is the center
r is the radius
In this problem we have
[tex](x-1)^{2}+(y-2)^{2}=3^{2}[/tex]
therefore
The coordinates of the center are [tex](h,k)=(1,2)[/tex]
The radius is equal to [tex]r=3\ units[/tex]
Part C) Plot the circle
using a graphing tool
see the attached figure
