Question

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

Answer 1

Answer:

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

To convert the given equation of the circle to standard form we should do as follows.

1. Add 36 to both sides of the equation.

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

2. We order the terms according to their variables.

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

3. Divide the linear term by 2, then elevate it to the square power.

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

4. Solve operations, and add the same units to the other side of the equation.

$x^{2} +4x+4+y^{2} -6y+9=36+4+9$

5. Now, we sum costant terms, and factor both trinomials.

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

As you can observe, the circle has center at (-2,3) and it has a radius of 7 units.

Notice that the second choice is correct, because we added 4 units to both sides to complete the square for the x terms.