Respuesta :
we have
[tex]x^{2} +y^{2} +4x-6y-36=0[/tex]
we know that
the equation of a circle in standard form is equal to
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
where
(h,k) is the center of the circle
r is the radius of the circle
Convert the equation to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
so
Adds [tex]36[/tex] both sides
[tex]x^{2} +y^{2} +4x-6y-36+36=0+36[/tex]
[tex]x^{2}+4x+y^{2}-6y=36[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side
[tex](x^{2}+4x+2^{2})+(y^{2}-6y+3^{2})=36+2^{2}+3^{2}[/tex]
[tex](x^{2}+4x+2^{2})+(y^{2}-6y+3^{2})=49[/tex]
Rewrite as perfect squares
[tex](x+2)^{2}+(y-3)^{2}=7^{2}[/tex]
the center of the circle is the point [tex](-2,3)[/tex]
the radius of the circle is [tex]7[/tex] units
Let's verify each of the statements
Statements
case A) To begin converting the equation to standard form, subtract [tex]36[/tex] from both sides
The statement is false
because, To begin converting the equation to standard form, adds [tex]36[/tex] from both sides
case B) To complete the square for the x terms, add [tex]4[/tex] to both sides
The statement is true
see the procedure
case C) The center of the circle is at [tex](-2,3)[/tex]
The statement is true
see the procedure
case D) The center of the circle is at [tex](4, -6)[/tex]
The statement is false
the center of the circle is the point [tex](-2,3)[/tex] (see the procedure)
case E) The radius of the circle is [tex]6[/tex]units
The statement is false
the radius of the circle is [tex]7[/tex] units (see the procedure)
case F) The radius of the circle is [tex]49[/tex] units
The statement is false
the radius of the circle is [tex]7[/tex] units (see the procedure)
therefore
the answer is
B. To complete the square for the x terms, add [tex]4[/tex] to both sides
C. The center of the circle is at [tex](-2,3)[/tex]
The center of the circle is (-2,3) and the radius of the circle is 7 units.
The general standard equation for the circle centered at ( h,k ) with radius r is,
[tex](x-h)^2 + (y-k)^2 =r^2[/tex]
Given-
The equation of the given circle is,
[tex]x^2+y^2+4x-6y-36=0[/tex]
To get the value of radius and center point solve it further to make it like a general equation.
Converting the equation to standard form
A- To begin converting the equation to standard form, substract 36 from both sides-
[tex]x^2+y^2+4x-6y-36=0[/tex]
[tex]x^2+y^2+4x-6y-36-36=0-36[/tex]
[tex]x^2+y^2+4x-6y=36[/tex]
B. To complete the square of the x terms add 4 to both sides,
[tex]x^2+y^2+4x-6y+4=36+4[/tex]
[tex]x^2+4x+4+y^2-6y=40[/tex]
Apply the algebraic identity used to find the square of the sum of two numbers, we get,
[tex](x+2)^2+y^2-6y=40[/tex]
To complete the square for the y terms add 9 to both sides,
[tex](x+2)^2+y^2-6y+9=40+9[/tex]
[tex](x+2)^2+(y-3)^2=49[/tex]
As by the above procedure both the A and B statements are true.
C- The center of the circle is at (-2,3)
Compare this equation with the general standard equation of a circle we get the center of the circle is (-2,3) and the radius of the circle is 7 units.
Hence statement C is true.
D- The center of the circle is at (4,-6).
As by the above solution, the center we get is (-2,3) and hence this is the false statement.
E- The radius of the circle is 6 units.
This is a false statement as the radius of the circle is 7 units.
F- The radius of the circle is 49 units.
This is a false statement as the radius of the circle is 7 units.
For more about the circle, follow the link below-
https://brainly.com/question/11833983