Question

Write the equation of a circle with center at (-4,12) and radius or square root 3

Answer 1

The general form of the equation of a circle is given as:

`(x-a)^2+(y-b)^2=r^2`

Where:

(a, b) is the coordinate of the center of the circle

r is the radius of the circle

Now, given the circle in question, with:

(-4, 12) as the coordinate of the center

sqrt(3) as the radius of the circle,

The equation of the circle is as follows:

`\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-(-4))^2+(y-12)^2=(\sqrt[]{3})^2 \\ (x+4)^2+(y-12)^2=(\sqrt[]{3})^2 \\ (x+4)^2+(y-12)^2=3^{} \end{gathered}`

Therefore, the equation of a circle with center (-4, 12) and radius sqrt(3) is:

(x + 4)^2 + (y - 12)^2 = 3