Another approach would be to use the Cartesian equation of the tangent through C to the circle and express that the distance from the center O of the circle to the tangent is equal to the radius. With axis X = AB and axis Y = D, center O = (1,1) , C = (4,4) Equation of the tangent passing thru C : ax + by + c = 0 with y-4 = m(x-4) ---> mx-y-4m+4 = 0 ---> a = m, b=1, c = -4m+4 Distance of O to tangent : abs(ax_O + by_O +c) / sqrt(a^2 + b^2) ....

Thank you so much for this great exlanation.

Another approach is to use angles. When you draw to perpendiculars from center to AB and CE, you get a deltoid. The arc inside the deltoid equals to angle CEB, and since the angle of deltoid's which is at the center of the circle sees that arc, it equals CEB. When you draw a line from center to E, you get an angle bisector so new angles are CEB/2. Then you can use tangent half angle formula in order to calculate x.

Let O be the center of the circle, let J and K be the points of tangency between circle O and DA and AB respectively, and let T be the point of tangency between circle O and EC. As all are radii of circle O, OJ = OK = OT = 1. DA and AB are tangent to circle O at J and K, so ∠OJA = ∠AKO = 90°. As ∠JAK = 90° also, then ∠KOJ = 90° and AKOJ is a square. EC is tangent to circle O at T, so ∠OTC = 90°. As ABCD and AKOJ are squares, both share vertex A, and AK and JA are coincident with AB and DA respectively, then O is colinear with AC. As AO, being the diagonal of a square, equals (1)√2 and AC = 4√2, OC = 3√2. Triangle ∆OTC: OT² + TC² = OC² 1³ + TC² = (3√2)² TC² = 18 - 1 = 17 TC = √17 Let ET = x. As KE and ET are tangents from circle O intersecting at E, KE = ET = x. Triangle ∆EBC: EB² + BC² = CE² (3-x)² + 4² = (√17+x)² 9 - 6x + x² + 16 = 17 + 2√17x + x² 2√17x + 6x = 25 - 17 (2√17+6)x = 8 x = 8/(2√17+6) x = 8(2√17-6)/(2√17+6)(2√17-6) x = 16(√17-3)/(68-36) x = 16(√17-3)/32 x = (√17-3)/2 EB = 3 - (√17-3)/2 EB = (6-√17+3)/2 EB = (9-√17)/2 A = ((9-√17)/2)(4)/2 = 9 - √17 sq units

such a good question and demonstrating

for every tangent there is a thales theorem: 10 dim x(3),y(3):l1=4:r=1:sw=.1:xm=r:ym=r:xs=xm:rem den thalessatz anwenden 20 xmt=(xm+l1)/2:ymt=(ym+l1)/2:rt=sqr((l1-xmt)^2+(l1-ymt)^2):goto 50 30 disy=rt^2-(xs-xmt)^2:if disy0 then 70 90 xs=(xs1+xs2)/2:gosub 30:if dg1*dg>0 then xs1=xs else xs2=xs 100 if abs(dg)>1E-10 then 90 110 print xs,"%",ys 120 dx=(l1-xs)/(l1-ys)*l1:xe=l1-dx:print xe:ages=(l1-xe)*l1/2:print "die flaeche=";ages 130 x(0)=0:y(0)=0:x(1)=l1:y(1)=0:x(2)=x(1):y(2)=l1:x(3)=0:y(3)=y(2) 140 mass=8E2/l1:goto 160 150 xbu=x*mass:ybu=y*mass:return 160 xba=0:yba=0:for a=1 to 4:ia=a:if ia=4 then ia=0 170 x=x(ia):y=y(ia):gosub 150:xbn=xbu:ybn=ybu:goto 190 180 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 190 gosub 180:next a:x=xm:y=ym:gcol5:gosub 150:circle xbu,ybu,r*mass 200 x=xe:y=0:gosub 150:xba=xbu:yba=ybu:x=l1:y=l1:gosub 150:xbn=xbu:ybn=ybu 210 gcol4:gosub 180 1.85385094% 0.479482396 1.56155281 die flaeche=4.87689437 > run in bbc basic sdl and hit ctrl tab to copy from the results window

9:35 can you explain where did the "x" in esquire route of 17 come from please?🤔

EO - биссектриса угла AEC, поэтому CE/AE = CO/AO = 3. Пусть z = BE

nice explanation

@MathBooster

6 ай бұрын

Thank you 😊

tan(2theta)=4/(3-x) is good as well as let angle NOE=theta

Eu fiquei em dúvida se a diagonal do quadrado passa no centro da circunferência.

@OverclockingCowboy

9 күн бұрын

You can draw 4 circles (radius=1) inside the square (side=4). The diagonal of the square will pass through the centers of 2 diagonal circles.

@OverclockingCowboy

9 күн бұрын

You can also divide the square (side=4) into 4 smaller squares (side=2).

this was on my year 8 maths test, except we were finding the hypotenuse, not the area. i remember no one knowing how to do this as the last question, and trying to do this for the next few nights. super traumatic.