What is the center of the circle x^2 + y^2 - 81 = 0? Simplify any fractions. (____,____)

Rewrite equation to standard form: We start by rewriting the equation to resemble the standard form of a circle's equation, which is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius. x^2 + y^2 - 81 = 0 Add 81 to both sides to isolate the x^2 and y^2 terms. x^2 + y^2 = 81Add 81 to both sides: Now, we compare the equation x^2 + y^2 = 81 with the standard form of a circle's equation. We notice that the x^2 and y^2 terms do not have any coefficients or constants added to them, which means that h and k are both 0. Therefore, the center of the circle is at (h,k) = (0,0).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

What is the center of the circle $x^2 + y^2 - 81 = 0$ ? $\newline$ Simplify any fractions. $\newline$ (,)

View step-by-step help

Home

Math Problems

Geometry

Find properties of circles from equations in general form

Full solution

Q. What is the center of the circle

x^2 + y^2 - 81 = 0

\newline

Simplify any fractions.

\newline

(____,____)

Rewrite equation to standard form: We start by rewriting the equation to resemble the standard form of a circle's equation, which is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center of the circle and $r$ is the radius. $\newline$ $x^2 + y^2 - 81 = 0$ $\newline$ Add $81$ to both sides to isolate the $x^2$ and $y^2$ terms. $\newline$ $x^2 + y^2 = 81$
Add $81$ to both sides: Now, we compare the equation $x^2 + y^2 = 81$ with the standard form of a circle's equation. We notice that the $x^2$ and $y^2$ terms do not have any coefficients or constants added to them, which means that $h$ and $k$ are both $0$ . Therefore, the center of the circle is at $(h,k) = (0,0)$ .