Calculate the remaining area of a quadrant with two non overlapping half circles

[Aussiebear]

Just trying to fill in some time ......

I have a quadrant with a radius of 6m, which contains two non overlapping half circles. One half circle with a radius of 3m adjoins the horizontal leg of the Quadrant, and the other half circle adjoins the vertical leg of the Quadrant, but starts at the top of the leg and we currently do not know its radius. I am trying to find the balance of the area remaining within the quadrant after calculating the Area of the two half circles.

Briefly speaking:
Step 1. Calculate the Area of the Quadrant

= (Pi()*6^2)/4)

Step 2. Calculate the area of the HC1 circle (the half circle adjoining the horizontal leg)

 =(Pi()*3^2)/2)

Step 3. Calculate the area of the HC2 circle (the half circle adjoining the vertical leg)
' This is where it gets a little messy. I created a triangle from the center point of the Quadrant to the mid point of HC1 to the mid point of HC2 and returning to the center point of the Quadrant. From this I calculated the length of the upright leg of the triangle and thus the radius of the HC2 circle.

Step 4. Area remaining = Area of Quadrant -(Sum(HC1 Area & Hc2 Area).

Providing i got that right (and its up for debate), I decided to push on and attempt to construct a Function.

Function CalculateQuadrantRemainingArea(QuadrantRadius As Double, HalfCircle1Radius As Double) As Double
    ' Calculates the remaining area of a quadrant with two non-overlapping half-circles.
    ' Half-circle 1 (HC1) is on the horizontal leg from the origin.
    ' Half-circle 2 (HC2) is on the vertical leg from the top of the quadrant.
    Const PI As Double = 3.14159265358979 
    Dim AreaQuadrant As Double
    Dim AreaHC1 As Double
    Dim RadiusHC2 As Double
    Dim AreaHC2 As Double
    Dim RemainingArea As Double
    ' Input validation (optional but recommended)
    If QuadrantRadius <= 0 Or HalfCircle1Radius <= 0 Then
        CalculateQuadrantRemainingArea = CVErr(xlErrValue) 
        ' Return #VALUE! error for invalid input
        Exit Function
    End If
    If HalfCircle1Radius > QuadrantRadius / 2 Then
        ' This scenario implies HC1 diameter is larger than the quadrant radius or it cannot fit fully on the leg from the origin.
        CalculateQuadrantRemainingArea = CVErr(xlErrValue) 
        ' Indicate an impossible scenario
        Exit Function
    End If
    ' Step 1: Calculate the area of the quadrant
    AreaQuadrant = (1 / 4) * PI * (QuadrantRadius ^ 2)
    ' Step 2: Calculate the area of the first half-circle (HC1)
    ' Assuming its diameter spans 2 * HalfCircle1Radius along the leg.
    AreaHC1 = (1 / 2) * PI * (HalfCircle1Radius ^ 2)
    ' Step 3: Determine the radius of the second half-circle (HC2)
    ' Using the non-overlapping (tangency) condition
    ' (r2^2 - 12r2 + 45 = 9 + 6r2 + r2^2) simplifies to 36 = 18r2 --> r2 = 2
    ' In the general case, where C1=(r1,0) and C2=(0,R-r2), we solve:
    ' Sqrt(r1^2 + (R-r2)^2) = r1 + r2
    ' r1^2 + (R-r2)^2 = (r1+r2)^2
    ' r1^2 + R^2 - 2*R*r2 + r2^2 = r1^2 + 2*r1*r2 + r2^2
    ' R^2 - 2*R*r2 = 2*r1*r2
    ' R^2 = 2*R*r2 + 2*r1*r2
    ' R^2 = r2 * (2R + 2r1)
    ' r2 = R^2 / (2 * (R + r1))
    RadiusHC2 = (QuadrantRadius ^ 2) / (2 * (QuadrantRadius + HalfCircle1Radius))
    ' Step 4: Calculate the area of the second half-circle (HC2)
    AreaHC2 = (1 / 2) * PI * (RadiusHC2 ^ 2)
    ' Step 5: Calculate the Remaining Area
    RemainingArea = AreaQuadrant - AreaHC1 - AreaHC2
    CalculateQuadrantRemainingArea = RemainingArea
End Function