Question

The equation of a circle whose center is at (4, 0) and radius is length 2√(3) is (x - 4)² + y² = 2√3 (x - 4)² + y² = 12 (x + 4)² + y² = 12

Answer 1

The equation of a circle is (x-a)²+(y-b)²=R²
(x-4)²+(y-0)²=(2√3)²
(x-4)²+y²=12

Answer 2

Answer:

Equation of the circle is

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

Step-by-step explanation:

We have been given that

center of the circle = (4,0)

Radius of the circle = $2\sqrt3$

The standard form of a circle is given by

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

Here, (h,k) is the center and r is the radius. Thus, we have

h = 4, k = 0, r = 2√(3)

Substituting these values in the above equation, we get

$(x-4)^2+(y-0)^2=(2\sqrt3)^2$

Simplifying, we get

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

Therefore, equation of the circle is

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