Question

Consider a circle whose equation is x2 + y2 + 4x – 6y – 36 = 0. Which statements are true? Check all that apply.

Answer 1

we have

$x^{2} +y^{2} +4x-6y-36=0$

we know that

the equation of a circle in standard form is equal to

$(x-h)^{2}+(y-k)^{2}=r^{2}$

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 $36$ both sides

$x^{2} +y^{2} +4x-6y-36+36=0+36$

$x^{2}+4x+y^{2}-6y=36$

Complete the square twice. Remember to balance the equation by adding the same constants to each side

$(x^{2}+4x+2^{2})+(y^{2}-6y+3^{2})=36+2^{2}+3^{2}$

$(x^{2}+4x+2^{2})+(y^{2}-6y+3^{2})=49$

Rewrite as perfect squares

$(x+2)^{2}+(y-3)^{2}=7^{2}$

the center of the circle is the point $(-2,3)$

the radius of the circle is $7$ units

Let's verify each of the statements

Statements

case A) To begin converting the equation to standard form, subtract $36$ from both sides

The statement is false

because, To begin converting the equation to standard form, adds $36$ from both sides

case B) To complete the square for the x terms, add $4$ to both sides

The statement is true

see the procedure

case C) The center of the circle is at $(-2,3)$

The statement is true

see the procedure

case D) The center of the circle is at $(4, -6)$

The statement is false

the center of the circle is the point $(-2,3)$ (see the procedure)

case E) The radius of the circle is $6$units

The statement is false

the radius of the circle is $7$ units (see the procedure)

case F) The radius of the circle is $49$ units

The statement is false

the radius of the circle is $7$ units (see the procedure)

therefore

the answer is

B. To complete the square for the x terms, add $4$ to both sides

C. The center of the circle is at $(-2,3)$

Answer 2

Complete the squaer for both
in
(x-h)^2+(y-k)^2=r^2
center is (h,k) and radius is r

x²+y²+4x-6y-36=0
group x and y terms
(x²+4x)+(y²-6y)-36=0
take 1/2 of linear cofients and squaer and add negative and positive inside
(x²+4x+4-4)+(y²-6y+9-9)-36=0
factor perfect square
((x+2)²-4)+((y-3)²-9)-36=0
distribute
(x+2)²-4+(y-3)²-9-36=0
add4+9+36 to both sides
(x+2)²+(y-3)²=49
(x-(-2))²+(y-3)²=7²

center is (-2,3) and radius is 7

answer is C
(not sure for A and B, because they could be, but they are not neccecary)