Question

a. (-3,4)
b. (3,-4)
c. (3,4)
d. (-3,-4)
what is the center of the circle given by (x + 3)^2 + (y - 4)^2 = 49?
a. complete the square
b. factor the equation
c. combine like terms
d. distribute terms
what transformation is needed to write the equation x^2 - y^2 - 6x + 4y + 9 = 0 in standard form?

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center of the circle.

Step2: Identify \(h\) and \(k\)

Given the equation \((x + 3)^2+(y - 4)^2 = 49\), we can rewrite \((x+3)\) as \((x-(-3))\). So, \(h=-3\) and \(k = 4\).

Step1: Analyze the equation \(x^{2}-y^{2}-6x + 4y+9 = 0\)

To write the equation of a conic (in this case, we can complete the square for \(x\) and \(y\) terms.
For the \(x\) - terms: \(x^{2}-6x=(x - 3)^{2}-9\)
For the \(y\) - terms: \(-y^{2}+4y=-(y^{2}-4y)=-(y - 2)^{2}+4\)

Step2: Complete the square process

$$
LATEXBLOCK0
$$

The first step in transforming \(x^{2}-y^{2}-6x + 4y+9 = 0\) to standard form is to complete the square for \(x\) and \(y\) terms.

Answer:

A. \((-3,4)\)