Question

Write the equation of a circle with center (-2, 0) and area = 64 pi

Answer 1

Hello!

The answer is:

The equation of the given circle is:

$(x+2)^{2} +(y)^{2}=64$

Why?

The equation of a circle is given by the following equation:

$(x-h)^{2} +(y-k)^{2}=r^{2}$

We are given the center point (-2,0) and the area of the circle.

The area of a circle is given by the formula:

$A=\pi*r^{2}\\64\pi=\pi*r^{2}\\64=r^{2}$

$A=\pi*r^{2}\\64\pi=\pi*r^{2}\\64=r^{2}\\\sqrt{64}=r\\8=r\\r=8$

So, the radius of the circle is 8 units.

Therefore,

We are given a circle where:

$h=x=-2\\k=y=0\\r=8$

Then, substituting into the circle equation, we have:

$(x-(-2))^{2} +(y-0)^{2}=(8)^$

$(x+2)^{2} +(y)^{2}=64$

Hence, the simplified equation of the circle is:

$(x+2)^{2} +(y)^{2}=64$

Have a nice day!

Answer 2

Answer:

The required equation in standard form is $(x+2)^2+y^2=64$

Step-by-step explanation:

The equation of a circle with center (h,k) an radius, r units is given by the formula;

$(x-h)^2+(y-k)^2=r^2$

The given circle has center (-2,0) and radius squared can be calculated from the given area, which is $64\pi$

$\pi r^2=64\pi$

$\implies r^2=64$

We substitute these values into the formula to obtain;

$(x--2)^2+(y-0)^2=64$

We simplify to get;

$(x+2)^2+y^2=64$