Question

Mrs. culland is finding the center of a circle whose equation is x2 y2 6x 4y – 3 = 0 by completing the square. her work is shown. x2 y2 6x 4y – 3 = 0 x2 6x y2 4y – 3 = 0 (x2 6x) (y2 4y) = 3 (x2 6x 9) (y2 4y 4) = 3 9 4 which completes the work correctly? (x – 3)2 (y – 2)2 = 42, so the center is (3, 2). (x 3)2 (y 2)2 = 42, so the center is (3, 2). (x – 3)2 (y – 2)2 = 42, so the center is (–3, –2). (x 3)2 (y 2)2 = 42, so the center is (–3, –2).

Answer 1

By completing squares, we will see that the center of the given circle is at (-3, -2).

How to find the center of the circle?

The general circle equation centered on (a, b) of radius R is:

$(x - a)^2 + (y - b)^2 = R^2$

In this case, our equation is:

$x^2 + y^2 + 6x + 4y - 3 = 0$

First, we need to complete squares, we will have:

$x^2 + 2*3*x + y^2 + 2*2*y - 3 = 0\\\\x^2 + 2*3*x + (9 - 9) + y^2 + 2*2*y + (4 - 4) - 3 = 0\\\\(x^2 + 2*3*x + 9) - 9 + (y^2 + 2*2*y + 4) - 4 - 3 = 0\\\\(x + 3)^2 + (y + 2)^2 = 3 + 9 + 4 = 16$

Then, by looking at the left side, we can see that the center of the circle is (-3, -2).

If you want to learn more about circles:

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Answer 2

Answer:

d

Step-by-step explanation:

got it right on edge