What is the equation of the circle C in which an equilateral triangle

Given, L1 x =0, line L2 3√y+x=3√3y+x=3 and Line L3 3√y−x+3√=03y−x+3=0
So, we can get the vertices value by the point of intersection of these lines.
Let triangle is ABC, so for getting A, B, C we should equate the lines.
=> L1 =L2 , L2 = L3, L3 = L1
So, upon solving we can get A (0,1) B (0,-1) and C(\(\sqrt{3}\),0)

A triangle is inscribed in the circle, then the distance from the center of the circle to all the vertices will be equal.

Let the Center of Circle is O(a,b).
Then using the distance formula:
AO = OB , OB = OC , OC = OA
Upon solving these equations we get a = (\(\sqrt{3}\)/3) and b= 0
For radius: The distance from the center to the Vertices will be the radius, as it`s equal.
So, radius = OA = OB = OC
upon solving any distance:
radius = 4/3

Thus, using circle formula: \((x-a)^2\) + \((y-b)^2\) = \((4/3)^2\)

Thus putting the values we get the answer as B.

Please hit the Kudos if you like the solution.

Intersection point of \(L_1\) and \(L_2\) is (0, 1) and that of \(L_1\) and \(L_3\) is (0, -1)

Center of the circle lies on perpendicular bisector of line \(L_1\)
Midpoint of Line \(L_1\) is (0, 0).
In center of the triangle= \(\frac{1}{2\sqrt{3}}\)*2= \(\frac{1}{\sqrt{3}}\)

Co-ordinates of center= (\(\frac{3}{\sqrt{3}}\), 0)

Radius of circle= \(\frac{1}{\sqrt{3}}\)*2=\(\frac{2}{\sqrt{3}}\)

[quote
So the circle will pass through (0, 1) and (0, -1).Only options (B) and (D) will satisfy these points.
The side of the equilateral triangle will be 2. Now note that the radius of the circle cannot be 2 since the side of the equilateral triangle must be greater than the radius.
So option (B) must be correct.[/quote]

VeritasKarishma can you please explain this further? how did option B & D satisfy 0,1 and 0, -1

Can you explain how did you calculate the diameter in D?

Please explain formulas used to calculate center and radius.