15:04 Once you've found PA, you can you can directly find PC using Pythagoras theorem (PBC is a right angled triangle). PC is the diameter as it projects a right angle on the circumference at B. PC /2 is the radius.

Let A(0,0), then B(6,0), C(6,-13), D(0,-18) O is the centre of the circle, then O(k,-13/2) OC=OD (k-6)^2+(13/2)^2=k^2+(23/2)^2 k=-9/2 radius=OC=sqrt((-9/2-6)^2+(13/2)^2)=sqrt(610)/2

Center at origin and B = (x,y) => x^^2 + y^^2 = r^^2 (x-6)^2 + (y-18)^2 = r^2 and x^2 + (y-13)^2 = r^2. Subtract the first to the 2 others and we get 2 linear equations with x and y. Solving gives x = 21/2 and y = 13/2 and this gives r.

Another way to do it is to draw the segment lines from O to D and B which are the radius of the circle. name x the distance from the center O to M. We consider the right triangles ONB and OMD. Observe that the length of the side BN is 13/2 and the length of MD is 18-13/2=23/2. Now use Pythagoras theorem for both triangle to write: R^2=(x+6)^2+(13/2)^2 and R^2=x^2+(23/2)^2. From this point no more geometry, now only algebra. It follows that (x+6)^2+(13/2)^2=x^2+(23/2)^2. This equation simplifies itself into 12x-54=0 or x=9/2. Therefore, R^2=(9/2)^2+(23/2)^2=81/4+529/4=610/4. Finally, R=sqrt(610)/2.

Extending BA to the left side of the circumference and a parallel chord from C also to the left circumference, we see there are 5 units below 13 units in the extended AD chord is 23 units, and the extended chord of AB is 21 units long. Center of the circle is at (21/2, 23/2) = (10.5, 11.5). Now R^2 - (11.5) ^2 + (10.5 - 6) ^2 = 11.5^2 + 4.5^2 = [23^2+9^2]/4 = (529+81)/4 = 610/4 --> R=sqrt (610) /2.

By symmetry, AF = 5, and by chords theorem AP= 15 and AB =6 given. Then form ∆ PFB, the radius is the product of diagonal ( PF * FB) divided by the double the height of ∆ (2*5) Hence,( √(15^2+5^2) * √(6^2+5^2))/(2*5) √250*√61/10 = 5*√610/10 = √610/2

I extended the 18 line to meet the circle, giving a line length 23 (symmetry). I constructed the perpendicular bisector of the 13 chord (which must pass through the centre of the circle and bisect the line length 23). If left intersect of perp to line 18 or 23 is "a" and from there to right intersect is "b" then we have two equations in a,b using the two chords length 23 and 13: a.b=23²/4 and (a+6)(b-6)=13²/4 which can be solved by substituting for b in the second eqn using the first eqn. Then a+b=2r so r=(a+b)/2

Let D=(0,0), then C=(6,5), B= (6,18) Let O be the center. Since the horizontal line from passing O bisects BC, we know O=(x, 11.5) Let M be the midpoint of C and D, we know M=(3, 2.5) Since OM is perpendicular to CD, we have (x-3,9) inner product with (6, 5)=0 6x-18+45 = 0, hence x=-4.5 Therefore, the radius r=\sqrt{(-4.5)^2+11.5^2}=(1/2)\sqrt{9^2+23^2}=sqrt{610}/2

I think it’s much simpler to use the formula for a circle. (x-a)^2 + (y-b)^2 = r^2 where a and b are the center coordinates of the circle. The intersection points are (0,0),(6,5)and(6,18) (arbitrary origin on bottom point). Then solve the three equations generated by the three intersection points. Easy!

(x-h)^2+(y-k)^2=r^2 B(x,y) = 0,0 C=0-13 D=-6,-18 then h^2+k^2=r^2 ...... eq1 h^2+-(13-k)^2=r^2 ...........eq2 (-6-h)^2+(-18-k)^2=r62 ......eq3 solve eq1=eq2 k=169/26 solve eq3=eq1 360-12h+h^2-36k+k^2=r^2 suitation k h=[360-(36*19/26)]/12 suitation h k eq1 r=12.349 As same answer

Las alineaciones BA y DA cortan la circunferencia en los puntos F y E respectivamente→ Potencia del punto A respecto a la circunferencia = AB*AF=AD*AE→ 6a=18*(18-13)→ a=15 → FC²=BF²+BC²→ (2r)²=(6+a)²+13²=21²+13²→ r=(√610)/2 =12,34908.... Gracias y un saludo cordial.

If I understand these methods, the first method triangulates to find the radius using the Pythagorean Thrm twice then uses a circle theorem and the second method requires applying chords twice. I thnk that this is the first time chords are diffcult to visualize.

Let O be the center of the circle. Draw radius OT, so that OT is perpendicular to BC and intersects it at M. As OT is perpendicular to BC and BC is parallel to AD, OT is also perpendicular to AD, and intersects it at N. As BC is a chord, and as OT is an intersecting radius that is perpendicular to BC, then OT bisects BC, thus BM = MC = 13/2. As ∠ANM = 90° and ∠NMB = 90°, then ABMN is a rectangle, and thus NM = AB = 6 and AN = BM = 13/2. As AD = 18, ND = 18-13/2 = 23/2. Extend AD from A to E so that ED is a chord. OT bisects ED for the same reasons as it does BC, therefore EN = ND. EN = ND EA + AN = ND EA + 13/2 = 23/2 EA = 10/2 = 5 By the intersecting chords theorem, if two chords of a circle intersect, then the products of the lengths of each side of the intersection are equal for each chord. Extend AB from A to F, so that FB is a chord. As A is the intersection point between FB and ED, then FA•AB = EA•AD. FA•AB = EA•AD 6FA = 5(18) = 90 FA = 90/6 = 15 Method 1: Draw OB, and draw radius OS such that OS is perpendicular to FB and intersects it at P. As above, OS bisects FB, so FP = PB = (15+6)/2 = 21/2. As PB is perpendicular to BC, PB = OM = 21/2. As BM is perpendicular to FB, OP = BM = 13/2. Triangle ∆OBM: BM² + OM² = OB² (13/2)² + (21/2)² = r² r² = 169/4 + 441/4 = 610/4 r = √(610/4) = √610/2 Method 2: If two chords intersect perpendicularly, then the sum of the squares of the divided chord segments equals 4 times the radius squared. AB² + FA² + AD² + EA² = 4r² 6² + 15² + 18² + 5² = 4r² 4r² = 36 + 225 + 324 + 25 = 610 r² = 610/4 r = √(610/4) = √610/2

after 15:00 :Use Pythagorean theorem on triangle PBC

At 13:39 PBC is an inscribed rectangle triangle then CP is a diameter; from 14:58 all you need is to use Pythagoras to find |CP|, and then R=|CP|/2: |CP|=√(21²+13²)=√(610), R= √(610)/2.

Second Method until the value of AP=15 is excellent.But then misses the obvious, that Inscribed Angle

Excelente solução! 🎉

Muito bom e muito obrigado!

La línea AD pasa por el centro del círculo?

Alternative "Insane" Method For number-crunching lovers only. If you're at all error prone with long complicated equations... don't even bother. If you like spending at least half an hour doing tons of algebra on a single problem... go for it. It took me three attempts, but it does work. Use chord bisector radius to create two Pythagoras equations. r² = (r - x)² + (13/2)² r² = [r - (x + 6)]² + [18 - (13/2)]² Have fun!

18-13=5 18*5=x*6 6x=90 x=15 (15+6)/2=21/2 (18+5)/2=23/2 R²=(21/2-6)²=(23/2)² R²=81/4+529/4=610/4 Radius of the circle = √610/2

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(18]^2= 244 (13)^2 =169 (6)^2 =36 {244A+169B+36C}= 419ABC 419ABC/360°=1.59ABC (ABC ➖ 59ABC+1)

Why not just use triangles OFM and OBN? We will have r^2=OM^2+MF^2 and r^2=ON^2+NB^2. NB=13/2=6.5, MF=MD=18-6.5=11.5, ON=OM+6. Let OM=x, then r^2=x^2+11.5^2, r^2=(x+6)^2+6.5^2. Subtracting first from second: 0=(x+6)^2+6.5^2-x^2-11.5^2=2*x*6+6^2+6.5^2-11.5^2;x=((11.5-6.5)(11.5+6.5)-36)/12=(5*18-36)/12=54/12=27/6. Then r=sqrt(27^2/6^2+11.5^2)=sqrt(152.5)=12.349.....

R^2=6,5^2+a^2...R^2=(18-6,5)^2+(a-6)^2...a=10,5...R^2=6,5^2+10,5^2=152,5

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2nd Method much better thanks

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EC in triangle DCB is not completely the height

би би

The solutions offered lack elegance. I much rather have my teeth extracted (without anesthetics) than attack this problem. What was the motivation in offering this problem?

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