Question

PLEASEEEEEEE PLEASEEEE HELPPPP ASAPPPPPP!! LOTS OF POINTS + BRAINLIEST :)

Find the error(s) and solve the problem correctly.
Find the equation of the circle. What is the center and radius of the circle?
Be sure to show all work and explain your answer.

X^2 + y^2 + 8x − 2y − 13 = 0
X^2 + 8x + y^2 − 2y = 13
X^2 + 8x + 8 + y^2 − 2y − 1 = 13 + 8 − 1
(x + 4)^2 + (y − 1)^2 = 20
Center (-4, 1) r = 20

Answer 1

Answer:

Center has coordinate $(-4; 1)$ and radius $\sqrt{30}$

Step-by-step explanation:

Let's start with rewriting terms as it was done:

$(x^2+8x) +(y^2-2y)-13=0$

Now let's focus on the two brackets: each one is one term from the square of a binomial, namely $(x+4)^2$ and $(y-1)^2$. By expanding them we get:

$x^2+8x+16$ and $y^2-2y+1$. Let's make the third term appear by adding and subtracting it (so we're not changing it) to our original equation:

$(x^2+8x+16)+(y^2-2y+1)-13-16-1=0$ at this point it's just crunching numbers and wrapping the squares back up

$(x+4)^2 +(y-1)^2 = 30$

At this point we can use the center-radius form of the equation of a circle ($(x-\alpha)^2 +(y-\beta)^2 = r^2$ to determine what we need by confronting the two

In particular, center has coordinate $(-4; 1)$ and radius $\sqrt{30}$