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

Mathematics

2 answers:

liberstina [14]3 years ago

7 0

You'll want to work with the center-radius form of a circle equation for this. the center formula is <span>(x – h)</span>²<span> + (y – k)</span>²<span> = r</span>², 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) = <span>y² - 24y + 144
</span>
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).

Dmitriy789 [7]3 years ago

4 0

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.

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