Question

What is the general form of the equation of the given circle with center A?

A: x2 + y2 + 6x − 24y − 25 = 0

B: x2 + y2 − 6x + 24y + 128 = 0

C: x2 + y2 + 6x – 24y + 128 = 0

D: x2 + y2 + 6x − 24y + 148 = 0

Answer 1

you'll want to work with the center-radius form of a circle equation for this. the center formula is (x – h)² + (y – k)² = r², where (h, k) is your center and r is your radius. plug in the information your circle gives you: A(-3, 12), radius = 5

(x + 3)² + (y - 12)² = (5)²  ... simplify the right side
(x + 3)² + (y - 12)² = 25 ... from here, you need to foil both of your binomials to convert this to the "general form" that your answer choices are in.
(x + 3)² = (x + 3)(x + 3) = x² + 6x + 9
(y - 12)² = (y - 12)(y - 12) = y² - 24y + 144

x² + 6x + 9 + y² - 24y + 144 = 25 ... combine like terms
x² + 6x + y² - 24y + 153 = 25 ... subtract 25
x² + 6x + y² - 24y + 128 = 0 is your equation. reorder it so that it's from the highest degree to the lowest: x² + y² + 6x - 24y + 128 = 0 is your result (C).

Answer 2

Theequation of circle with center at $\left( { - 3,12} \right)$ and radius $5{\text{ units}}$ is $\boxed{{x^2} + {y^2} + 6x - 24y + 128 = 0}.$

Further explanation:

The standard equation of the circle with center $\left( {h,k} \right)$ and radius r can be expressed as,

${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}.$

Given:

A circle with center at $\left( { - 3,12} \right).$

Radius of the circle is $5{\text{ units}}.$

Explanation:

The center is at $\left( { - 3,12}\right).$

Compare the general form $\left( {h,k} \right)$ of the center to the given center $\left( { - 3,12} \right)$ to obtain the value of $h$ and $k$.

Therefore, the value of $h$ and $k$ are $-3$ and $12$ respectively.

The radius of the circle is $5{\text{ units}}.$

Substitute $- 3$ for $h$, $12$ for $k$ and $5$ for $r$ to obtain the standard equation of the circle.

$\begin{aligned}{\left( {x - \left( { - 3} \right)} \right)^2} + {\left({y - 12} \right)^2}&= {\left( 5 \right)^2}\\{\left( {x + 3} \right)^2} + {\left( {y - 12}\right)^2} &= 25\\{x^2} + 6x + 9 + {y^2} - 24y + 144&= 25\\{x^2} + {y^2} + 6x - 24y + 153&= 25\\{x^2} + {y^2} + 6x - 24y + 153 - 25&= 0\\{x^2} + {y^2} + 6x - 24y + 128&= 0\\\end{aligned}$

The equation of circle with center at $\left( { - 3,12}\right)$ and radius $5{\text{ units}}$ is $\boxed{{x^2} + {y^2} + 6x - 24y + 128 = 0}.$

Learn more:

1. Learn more about equation of circle brainly.com/question/1506955.

2. Learn more about domain of the function https://brainly.com/question/3852778.

3. Learn more about coplanar https://brainly.com/question/4165000.

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Circle

Keywords: Circle, standard form of the circle, equation of the circle, center, diameter of circle, radius of the circle, center A.