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what is the center of the circle (x + 1/4)^2 + y^2 = 64? simplify any f…

Question

what is the center of the circle (x + 1/4)^2 + y^2 = 64? simplify any fractions.

Explanation:

Step1: Recall the standard - form of a circle equation

The standard - form of a circle equation is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center of the circle and $r$ is the radius.

Step2: Rewrite the given equation in the standard form

The given equation is $(x+\frac{1}{4})^2 + y^2=64$, which can be written as $(x-(-\frac{1}{4}))^2+(y - 0)^2 = 8^2$.

Answer:

$(-\frac{1}{4},0)$