What is the center of the circle x2+y2–144=0? Simplify any fractions. (□,□)

Write given equation: We start by writing the given equation of the circle: x^2 + y^2 – 144 = 0. To find the center, we need to get the equation in the standard form of a circle, which is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius.Move constant term: We move the constant term to the other side of the equation by adding 144 to both sides: x^2 + y^2 – 144 + 144 = 0 + 144. This simplifies to x^2 + y^2 = 144.Compare to standard form: Now, we compare the equation x^2 + y^2 = 144 to the standard form (x-h)^2 + (y-k)^2 = r^2. Since there are no (x-h) or (y-k) terms, it means that h and k are both 0.Find center: Therefore, the center of the circle is at (h, k) = (0, 0).

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What is the center of the circle $x^2+y^2-144=0$ ? $\newline$ Simplify any fractions. $\newline$ $(\square, \square)$

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Find properties of circles from equations in general form

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Q. What is the center of the circle

x^2+y^2-144=0

\newline

Simplify any fractions.

\newline

(\square, \square)

Write given equation: We start by writing the given equation of the circle: $x^2 + y^2 - 144 = 0$ . To find the center, we need to get the equation in the standard form of a circle, which is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center and $r$ is the radius.
Move constant term: We move the constant term to the other side of the equation by adding $144$ to both sides: $x^2 + y^2 - 144 + 144 = 0 + 144$ . This simplifies to $x^2 + y^2 = 144$ .
Compare to standard form: Now, we compare the equation $x^2 + y^2 = 144$ to the standard form $(x-h)^2 + (y-k)^2 = r^2$ . Since there are no $(x-h)$ or $(y-k)$ terms, it means that $h$ and $k$ are both $0$ .
Find center: Therefore, the center of the circle is at $(h, k) = (0, 0)$ .