RetroBASIC

My memory of BBC Basic was that is was rather complex process to get Mouse X, Y but that was some time ago.

BBC for Windows was OK too.

Great learning environment for kids making use of the Logo programming language

https://www.turtleacademy.com/
 
Its great fun 

Here is SmallBASIC version of Knot #3, almost identical to QB64:
'"Celtic Knot 3" ' B+ 2019-07-12 from QB64
' To demo Celtic Knot mastery, attempt another one!

borderColor = RGB(255, 255, 0)
fillColor = RGB(0, 128, 0)
moreColor = RGB(160, 160, 255)
FOR rr = 30 TO 350 STEP 20
    CLS
    Knot3 xmax / 2, ymax / 2, rr, .12 * rr, borderColor, fillColor, moreColor
    CIRCLE xmax / 2, ymax / 2, rr, 1, rgb(128, 128, 128)
    INPUT "OK ..."; w$
NEXT

'KnotR tested OK from 30 to 350, thick 1 to 13% knotR, at thick = 1 and 2 just see fillC.
SUB Knot3 (xc, yc, knotR, thick, borderC, fillC, middleC)
    'DIM p, r, s, t, br, fr, x, y
    pm2d3 = pi * 2 / 3 '  all crucial points are 1/3 circle symmetric
    r = knotR / 2.6: p = 35 * pi * r: s = 2 * pi / p
    br = thick / 2 'border radius
    fr = br - 4
    mr = fr - 4

    'outline whole design
    FOR t = 0 TO 2 * PI STEP s
        x = xc + r * (COS(t) + COS(4 * t) / .7 + SIN(2 * t) / 12)
        y = yc + r * (SIN(t) + SIN(4 * t) / .7 + COS(2 * t) / 12)
        'PSET (x, y)
        fcirc x, y, br, borderC
        'code to help me find where in the 0 to 2*PI points are
        'IF t = 0 THEN fcirc x, y, 3, &HFF0000FF
        'IF ABS(t - pi) < .0025 THEN fcirc x, y, 3, &HFF000066
    NEXT
    'fill in design, same as above only smaller circle fills
    FOR t = 0 TO 2 * PI STEP s
        x = xc + r * (COS(t) + COS(4 * t) / .7 + SIN(2 * t) / 12)
        y = yc + r * (SIN(t) + SIN(4 * t) / .7 + COS(2 * t) / 12)
        'PSET (x, y)
        fcirc x, y, fr, fillC
    NEXT
    FOR t = 0 TO 2 * PI STEP s
        x = xc + r * (COS(t) + COS(4 * t) / .7 + SIN(2 * t) / 12)
        y = yc + r * (SIN(t) + SIN(4 * t) / .7 + COS(2 * t) / 12)
        'PSET (x, y)
        fcirc x, y, mr, middleC
    NEXT
    'over bridges  borders, locate over the top passes and draw circles over the over pass
    FOR t = 0 TO 2 * PI STEP s
        x = xc + r * (COS(t) + COS(4 * t) / .7 + SIN(2 * t) / 12)
        y = yc + r * (SIN(t) + SIN(4 * t) / .7 + COS(2 * t) / 12)

        IF ABS(t - pi * 51 / 384) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 51 / 384 - pm2d3) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 51 / 384 - 2 * pm2d3) < .11 THEN fcirc x, y, br, borderC

        IF ABS(t - pi * 111 / 384) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 111 / 384 - pm2d3) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 111 / 384 - 2 * pm2d3) < .11 THEN fcirc x, y, br, borderC


        IF ABS(t - pi * 424 / 384) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 424 / 384 - pm2d3) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 424 / 384 + 1 * pm2d3) < .11 THEN fcirc x, y, br, borderC

        IF ABS(t - pi * 680 / 384) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 680 / 384 + 2 * pm2d3) < .11 THEN fcirc x, y, br, borderC
        IF ABS(t - pi * 680 / 384 + 1 * pm2d3) < .11 THEN fcirc x, y, br, borderC
    NEXT
    'over bridges fills , now draw farther up and down the bidge work the fill color
    FOR t = 0 TO 2 * PI STEP s
        x = xc + r * (COS(t) + COS(4 * t) / .7 + SIN(2 * t) / 12)
        y = yc + r * (SIN(t) + SIN(4 * t) / .7 + COS(2 * t) / 12)

        IF ABS(t - pi * 51 / 384) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 51 / 384 - pm2d3) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 51 / 384 - 2 * pm2d3) < .13 THEN fcirc x, y, fr, fillC

        IF ABS(t - pi * 111 / 384) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 111 / 384 - pm2d3) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 111 / 384 - 2 * pm2d3) < .13 THEN fcirc x, y, fr, fillC

        IF ABS(t - pi * 424 / 384) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 424 / 384 - pm2d3) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 424 / 384 + 1 * pm2d3) < .13 THEN fcirc x, y, fr, fillC

        IF ABS(t - pi * 680 / 384) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 680 / 384 + 2 * pm2d3) < .13 THEN fcirc x, y, fr, fillC
        IF ABS(t - pi * 680 / 384 + 1 * pm2d3) < .13 THEN fcirc x, y, fr, fillC
    NEXT
    FOR t = 0 TO 2 * PI STEP s
        x = xc + r * (COS(t) + COS(4 * t) / .7 + SIN(2 * t) / 12)
        y = yc + r * (SIN(t) + SIN(4 * t) / .7 + COS(2 * t) / 12)

        IF ABS(t - pi * 51 / 384) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 51 / 384 - pm2d3) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 51 / 384 - 2 * pm2d3) < .15 THEN fcirc x, y, mr, middleC

        IF ABS(t - pi * 111 / 384) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 111 / 384 - pm2d3) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 111 / 384 - 2 * pm2d3) < .15 THEN fcirc x, y, mr, middleC
       
        IF ABS(t - pi * 424 / 384) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 424 / 384 - pm2d3) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 424 / 384 + 1 * pm2d3) < .15 THEN fcirc x, y, mr, middleC

        IF ABS(t - pi * 680 / 384) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 680 / 384 + 2 * pm2d3) < .15 THEN fcirc x, y, mr, middleC
        IF ABS(t - pi * 680 / 384 + 1 * pm2d3) < .15 THEN fcirc x, y, mr, middleC
    NEXT
END SUB

SUB fcirc (cX, cY, R, C)
  circle cx, cy, r, 1, C filled
END SUB


A much easier approach, you don't have to eye ball to the pixel the start and end of arcs nor the heavy math to get 2 circle intersect points. This approach lays out the design shape and then draws "bridges" where the under/over passes are. Probably closer to how a human artist might do a Celtic Knot. Also don't worry about the inner and outer circle radii, just one radius for center of rings.

OPTION _EXPLICIT
_TITLE "Celtic Knot 2" ' B+ developed in JB, translated to QB64 2019-07-11
' Instead of worrying about 2 circles (or arcs of them) for one ring, draw one circle with really wide pen!
' Draw shape twice with Outline color then Fill color (smaller width) then lay "Bridges"
' over sections for underpass/overpass. Thanks to tsh73 for demo of this method with another Celtic Knot.

CONST xmax = 720, ymax = 720, pi = 3.14159265, xc = xmax / 2, yc = ymax / 2
SCREEN _NEWIMAGE(xmax, ymax, 32)
_SCREENMOVE 100, 20
COLOR , &HFF888888: CLS
DIM r
r = 350
knot2 xc, yc, r, 30, &HFFFFFF00, &HFF008800
CIRCLE (xc, yc), r
SLEEP

'for best results: 50+ (pixels) and up for knotR to see details
'max thickness for knotR (impreceise overall radius) is about 10%
SUB knot2 (x, y, knotR, thick, outlineC AS _UNSIGNED LONG, fillC AS _UNSIGNED LONG)
    DIM hCenterX(6), hCenterY(6), hCX(6), hCY(6)
    DIM w, hexAngle, h, hexD2, PD96, tD2
    hexAngle = pi / 3: hexD2 = hexAngle / 2: PD96 = pi / 96 '<<<< all angles are refined to 96 parts of semicircle
    w = knotR / 15: tD2 = thick / 2
    FOR h = 1 TO 6
        hCenterX(h) = x + 8 * w * COS(h * hexAngle + pi / 6)
        hCenterY(h) = y + 8 * w * SIN(h * hexAngle + pi / 6)
        hCX(h) = x + 14 * w * COS(h * hexAngle)
        hCY(h) = y + 14 * w * SIN(h * hexAngle)
    NEXT
    FOR h = 1 TO 6
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + pi / 6 + PD96 * 7, h * hexAngle + pi / 6 + PD96 * 185, tD2, outlineC
        penArc hCX(h), hCY(h), 6 * w, h * hexAngle + PD96 * 55, h * hexAngle + PD96 * 137, tD2, outlineC
    NEXT
    FOR h = 1 TO 6
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + pi / 6 + PD96 * 7, h * hexAngle + pi / 6 + PD96 * 185, tD2 - 1, fillC
        penArc hCX(h), hCY(h), 6 * w, h * hexAngle + PD96 * 55, h * hexAngle + PD96 * 137, tD2 - 1, fillC
    NEXT
    'bridges
    FOR h = 1 TO 6
        'boarder color arc 7
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + PD96 * 51, h * hexAngle + PD96 * 58, tD2, outlineC
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + PD96 * 102, h * hexAngle + PD96 * 109, tD2, outlineC
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + PD96 * 148, h * hexAngle + PD96 * 155, tD2, outlineC
        penArc hCX(h), hCY(h), 6 * w, h * hexAngle + PD96 * 83, h * hexAngle + PD96 * 90, tD2, outlineC

        'fill color arc 2 before 3 after
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + PD96 * 49, h * hexAngle + PD96 * 61, tD2 - 1, fillC
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + PD96 * 99, h * hexAngle + PD96 * 111, tD2 - 1, fillC
        penArc hCenterX(h), hCenterY(h), 6 * w, h * hexAngle + PD96 * 145, h * hexAngle + PD96 * 158, tD2 - 1, fillC
        penArc hCX(h), hCY(h), 6 * w, h * hexAngle + PD96 * 80, h * hexAngle + PD96 * 92, tD2 - 1, fillC
    NEXT
END SUB

SUB penArc (x, y, r, raStart, raStop, penWidth, c AS _UNSIGNED LONG)
    'x, y origin, r = radius, c = color
    'raStart is first angle clockwise from due East = 0 degrees
    ' arc will start drawing there and clockwise until raStop angle reached
    DIM aStep, a
    IF raStop < raStart THEN
        penArc x, y, r, raStart, pi * 2, penWidth, c
        penArc x, y, r, 0, raStop, penWidth, c
    ELSE
        aStep = 1 / (pi * r * 2)
        FOR a = raStart TO raStop STEP aStep
            fcirc x + r * COS(a), y + r * SIN(a), penWidth, c
        NEXT
    END IF
END SUB

'no built in circle fill sub with QB64 but this one is good!
SUB fcirc (CX AS INTEGER, CY AS INTEGER, R AS INTEGER, C AS _UNSIGNED LONG)
    DIM Radius AS INTEGER, RadiusError AS INTEGER
    DIM X AS INTEGER, Y AS INTEGER
    Radius = ABS(R): RadiusError = -Radius: X = Radius: Y = 0
    IF Radius = 0 THEN PSET (CX, CY), C: EXIT SUB
    LINE (CX - X, CY)-(CX + X, CY), C, BF
    WHILE X > Y
        RadiusError = RadiusError + Y * 2 + 1
        IF RadiusError >= 0 THEN
            IF X <> Y + 1 THEN
                LINE (CX - Y, CY - X)-(CX + Y, CY - X), C, BF
                LINE (CX - Y, CY + X)-(CX + Y, CY + X), C, BF
            END IF
            X = X - 1
            RadiusError = RadiusError - X * 2
        END IF
        Y = Y + 1
        LINE (CX - X, CY - Y)-(CX + X, CY - Y), C, BF
        LINE (CX - X, CY + Y)-(CX + X, CY + Y), C, BF
    WEND
END SUB


The corners are nice and round, more natural:

OK here is code to screen shot at start of this thread:
OPTION _EXPLICIT
_TITLE "Vince Celtic Challenge Skeleton"
CONST xmax = 1200, ymax = 700, pi = 3.14159265
CONST xc = xmax / 2, yc = ymax / 2
SCREEN _NEWIMAGE(xmax, ymax, 32)
_SCREENMOVE 100, 20

celticKnot xc, yc, 30
SLEEP

SUB celticKnot (xc, yc, thick)
    DIM distr, r, r1, hex, hd2, nHex, x0, y0, c AS _UNSIGNED LONG

    distr = 5 * thick
    r = 4 * thick
    r1 = 3 * thick
    hex = pi * 2 / 6
    hd2 = hex / 2

    FOR nHex = 1 TO 6
        IF nHex <= 2 THEN c = &HFFFFFFFF ELSE c = &HFF444444
        'circles with origins in main circle
        x0 = xc + distr * COS(nHex * hex - hd2)
        y0 = yc + distr * SIN(nHex * hex - hd2)

        'outer radius
        CIRCLE (x0, y0), r, c
        'inner radius
        CIRCLE (x0, y0), r1, c


        ' circles with outside origins
        x0 = xc + (2 * distr - 1.5 * thick) * COS(nHex * hex)
        y0 = yc + (2 * distr - 1.5 * thick) * SIN(nHex * hex)
        IF nHex = 1 THEN c = &HFF0000FF ELSE c = &HFF444444
        'outer radius
        CIRCLE (x0, y0), r, c
        'inner radius
        CIRCLE (x0, y0), r1, c

    NEXT
END SUB