We can use intersecting secants theorem for secants to the yellow circle originating from O: 2*(2 + r - 2 + r - 2) = 14*(14 - 2r) which gives r = 25/4 readily.
Let's find the area: . .. ... .... ..... First of all we calculate the radius R of the quarter circle: A = πR²/4 49π = πR²/4 4*49 = R² ⇒ R = √(4*49) = 2*7 = 14 The circumference of the quarter circle and the circumference of the yellow circle have exactly one point of intersection. Therefore this point and the centers of the two circles are located on the same line. So with r being the radius of the yellow circle we obtain: OC + BC = OB OC + r = R ⇒ OC = R − r A is a point of tangency. Therefore AC is parallel to OQ and the triangle OAC is a right triangle. By applying the Pythagorean theorem we obtain: OA² + AC² = OC² OA² + r² = (R − r)² ⇒ OA² = (R − r)² − r² Now let's add point R such that AR is the diameter of the yellow circle. According to the theorem of Thales the triangle APR is a right triangle. Let's also add point S such that PS is the height of the triangle APR according to its base AR. Since APR is a right triangle, we can apply the right triangle altitude theorem: AS*RS = PS² AS*(AR − AS) = PS² OP*(2*r − OP) = OA² OP*(2*r − OP) = (R − r)² − r² 2*(2*r − 2) = (14 − r)² − r² 4*r − 4 = 196 − 28*r + r² − r² 32*r = 200 ⇒ r = 200/32 = 25/4 Now we are able to calculate the area of the yellow circle: A(yellow circle) = πr² = (625/16)π ≈ 122.72 Best regards from Germany
Join OCB( these points are collinear according to tangent / kissing circle property) . Suppose it cuts yellow circle at point R . In respect of yellow circle OB & OQ are secants. Hence by intersecting secants theorem we may get OP*OQ = OR * OB---(1) OP=2 (given) OQ = 2+2(r -2) OR= R (radius of the quarter circle) = 14 OR= OB - BOR =14 -2r Hence from equation (1) we may write 2[2+2(r - 2) = (14 - 2r) 14 > 4r - 4 = 196 - 28r > r = 200/32=25/4 Area of yellow circle = 22/7*25/4*25/4 =122 . 77 sq units (approx.)
In triangle OCA Hypotenuse (OC) =14 - r (circles touches internally and distance between the centres is difference of their radii) (14 - r) ^2 = r^2 + [2√(r - 1)]^2 >196 - 28r +r^2 =r^2+4r -4 > r = 200/32
After figuring out that the radius of the quarter circle is 14, you can solve for the radius of the yellow circle with 2 right triangles: 1) OA^2+r^2=(14-r)^2 and 2) OA^2+(r-2)^2=r^2. With triangle 1, you get OA^2=196-28r. With triangle 2, you get OA^2= 4r-4. Setting these two equations equal to each other results in: 4r-4 = 196-28r. Solving for r, we get r= 25/4.
Let's use an orthonormal center O, first axis (OA) The equation of the yellow circle is: x^2 + y^2 +a.x + b.y +c = 0 P(0; 2) is on this circle, so 4 + 2.b + c = 0 and c = -2.b -4 The equation is now x^2 + y^2 +a.x +b.y -2.b -4 = 0 At the intersection with (OA) (which equation is y = 0) we have: x^2 + a.x -2.b -4 = 0. This equation must have a double solution (point of tangency) so its delta is 0 Delta = a^2 +4.(2.b +4) = a^2 + 8.b +16 = 0 So we have b = (-(a^2)/8) -2 Now the equation of the yellow circle is: x^2 + y^2 +a.x - ((a^2 +16)/8).y +(a^2)/4 = 0 or (x +a/2)^2 + (y - (a^2 +16)/16)^2 = ((a^2 + 16)/16)^2 The radius of the yellox circle is then r = (a^2 +16)/16 (*) The center C of this circle has for abscissa -a/2, so OA = -a/2 and OC^2 = ((a^2) /4 ) + r^2 in OAC Then OC = sqrt((a^2)/4) + r^2) OC + CB = radius of the big quater circle = 14 So we have:sqrt((a^2)/4 +r^2) + r =14 (a^2)/4 + r^2 = (14 - r)^2. , then (a^2)/4 = -28.(r^2) + 196 We replace (a^2)/4 by 4.(r^2) -4 (by *) and have: 32.r = 200 and r = 200/32 = 25/4 The yellow area is Pi.((25/4)^2) = (625/16).Pi Another method to find r at the end: When the coordinates of C are known, we can write the equation of (OC), find the intersection B with the big quater circle and then r = CB. (This problem was not so evident.)
Let R be the radius of quarter circle O and r be the radius of circle C. Quarter circle O: Aₒ = πR²/4 49π = πR²/4 R² = 49π(4/π) = 196 R = √196 = 14 Draw radius OB. When two circles are tangent to each other their point of tangency is collinear with their centers. As B is the point of tangency between circle C and quarter circle O, then C is collinear with OB. As CB = r and OB = R = 14, OC = 14-r. Draw CA. As CA is a radius and OA is tangent to circle C at A, ∠OAC = 90°. Triangle ∆OAC: OA² + CA² = OC² OA² + r² = (14-r)² OA² = 196 - 28r + r² - r² OA = √(196-28r) Draw CM, where M is the point on PQ where CM is perpendicular to PQ. As PQ is a chord of circle C, CM bisects PQ, so PM = MQ. As ∠CMP = ∠MOA = ∠OAC = 90°, then ∠ACM = 90° and OACM is a rectangle. As CA = OM = r and OP = 2, PM = r-2. Triangle ∆CMP: CM² + PM² = CP² (196-28r) + (r-2)² = r² 196 - 28r + r² - 4r + 4 = r² 32r = 200 r = 200/32 = 25/4 Circle C: Aᴄ = πr² = π(25/4)² = 625π/16 ≈ 122.72 sq units
They are both tangent at point B. So by def, the line from their center to B is perpendicular (90 angle) to the tangent line. If they both have a perpendicular line at B going to their center, then the lines are colinear, and OCB is a straight line.
Fantastique
Merci chéri❤️
We can use intersecting secants theorem for secants to the yellow circle originating from O:
2*(2 + r - 2 + r - 2) = 14*(14 - 2r)
which gives r = 25/4 readily.
The simplest solution. Well done.
I did it the same way!😊
Thanks for the feedback ❤️
Thank you! Good introduction to the tangent secant theorem.
Glad you enjoyed it!
Thanks for the feedback ❤️
Let's find the area:
.
..
...
....
.....
First of all we calculate the radius R of the quarter circle:
A = πR²/4
49π = πR²/4
4*49 = R²
⇒ R = √(4*49) = 2*7 = 14
The circumference of the quarter circle and the circumference of the yellow circle have exactly one point of intersection. Therefore this point and the centers of the two circles are located on the same line. So with r being the radius of the yellow circle we obtain:
OC + BC = OB
OC + r = R
⇒ OC = R − r
A is a point of tangency. Therefore AC is parallel to OQ and the triangle OAC is a right triangle. By applying the Pythagorean theorem we obtain:
OA² + AC² = OC²
OA² + r² = (R − r)²
⇒ OA² = (R − r)² − r²
Now let's add point R such that AR is the diameter of the yellow circle. According to the theorem of Thales the triangle APR is a right triangle. Let's also add point S such that PS is the height of the triangle APR according to its base AR. Since APR is a right triangle, we can apply the right triangle altitude theorem:
AS*RS = PS²
AS*(AR − AS) = PS²
OP*(2*r − OP) = OA²
OP*(2*r − OP) = (R − r)² − r²
2*(2*r − 2) = (14 − r)² − r²
4*r − 4 = 196 − 28*r + r² − r²
32*r = 200
⇒ r = 200/32 = 25/4
Now we are able to calculate the area of the yellow circle:
A(yellow circle) = πr² = (625/16)π ≈ 122.72
Best regards from Germany
Excellent!
Thanks for sharing ❤️
Join OCB( these points are collinear according to tangent / kissing circle property)
. Suppose it cuts yellow circle at point R .
In respect of yellow circle
OB & OQ are secants.
Hence by intersecting secants theorem we may get
OP*OQ = OR * OB---(1)
OP=2 (given)
OQ = 2+2(r -2)
OR= R (radius of the quarter circle) = 14
OR= OB - BOR =14 -2r
Hence from equation (1) we may write
2[2+2(r - 2) = (14 - 2r) 14
> 4r - 4 = 196 - 28r
> r = 200/32=25/4
Area of yellow circle
= 22/7*25/4*25/4
=122 . 77 sq units (approx.)
Excellent!
Thanks for sharing ❤️
intresting
Glad to hear that!
Thanks for the feedback ❤️
In triangle OCA
Hypotenuse (OC) =14 - r (circles touches internally and distance between the centres is difference of their radii)
(14 - r) ^2 = r^2 + [2√(r - 1)]^2
>196 - 28r +r^2 =r^2+4r -4
> r = 200/32
Thanks for the feedback ❤️
Nice ... Congrats
Glad to hear that!
Thanks for the feedback ❤️
After figuring out that the radius of the quarter circle is 14, you can solve for the radius of the yellow circle with 2 right triangles: 1) OA^2+r^2=(14-r)^2 and 2) OA^2+(r-2)^2=r^2. With triangle 1, you get OA^2=196-28r. With triangle 2, you get OA^2= 4r-4. Setting these two equations equal to each other results in: 4r-4 = 196-28r. Solving for r, we get r= 25/4.
Excellent!
Thanks for sharing ❤️
This is really awesome,many thanks, Sir, very challenging 🙂
fast lane: AO = 2√(r - 1); BO = 14 = BT + TO = 2r + (14 - 2r) → 4(r - 1) = 196 - 28r → r = 25/4
btw: AC = AN + = r + (r - 2) → CPN = α → sin(α) = 17/25; COA = δ → sin(δ) = 25/31
You are very welcome!
Thanks for the feedback ❤️
OA=a..raggio(big)=14...le equazioni sono R+√(R^2+a^2)=14..R=2+√(R^2-a^2)... risulta a^2=4R-4..e la sostituisco nella prima..32R=200...R=25/4
Excellent!
Thanks for sharing ❤️
Let's use an orthonormal center O, first axis (OA)
The equation of the yellow circle is:
x^2 + y^2 +a.x + b.y +c = 0
P(0; 2) is on this circle, so 4 + 2.b + c = 0 and c = -2.b -4
The equation is now x^2 + y^2 +a.x +b.y -2.b -4 = 0
At the intersection with (OA) (which equation is y = 0) we
have: x^2 + a.x -2.b -4 = 0. This equation must have a double solution (point of tangency) so its delta is 0
Delta = a^2 +4.(2.b +4) = a^2 + 8.b +16 = 0
So we have b = (-(a^2)/8) -2
Now the equation of the yellow circle is:
x^2 + y^2 +a.x - ((a^2 +16)/8).y +(a^2)/4 = 0
or (x +a/2)^2 + (y - (a^2 +16)/16)^2 = ((a^2 + 16)/16)^2
The radius of the yellox circle is then r = (a^2 +16)/16 (*)
The center C of this circle has for abscissa -a/2,
so OA = -a/2 and OC^2 = ((a^2) /4 ) + r^2 in OAC
Then OC = sqrt((a^2)/4) + r^2)
OC + CB = radius of the big quater circle = 14
So we have:sqrt((a^2)/4 +r^2) + r =14
(a^2)/4 + r^2 = (14 - r)^2. , then (a^2)/4 = -28.(r^2) + 196
We replace (a^2)/4 by 4.(r^2) -4 (by *) and have:
32.r = 200 and r = 200/32 = 25/4
The yellow area is Pi.((25/4)^2) = (625/16).Pi
Another method to find r at the end: When the coordinates of C are known, we can write the equation of (OC), find the intersection B with the big quater circle and then r = CB.
(This problem was not so evident.)
Excellent!
Thanks for the feedback ❤️
Let R be the radius of quarter circle O and r be the radius of circle C.
Quarter circle O:
Aₒ = πR²/4
49π = πR²/4
R² = 49π(4/π) = 196
R = √196 = 14
Draw radius OB. When two circles are tangent to each other their point of tangency is collinear with their centers. As B is the point of tangency between circle C and quarter circle O, then C is collinear with OB. As CB = r and OB = R = 14, OC = 14-r. Draw CA. As CA is a radius and OA is tangent to circle C at A, ∠OAC = 90°.
Triangle ∆OAC:
OA² + CA² = OC²
OA² + r² = (14-r)²
OA² = 196 - 28r + r² - r²
OA = √(196-28r)
Draw CM, where M is the point on PQ where CM is perpendicular to PQ. As PQ is a chord of circle C, CM bisects PQ, so PM = MQ. As ∠CMP = ∠MOA = ∠OAC = 90°, then ∠ACM = 90° and OACM is a rectangle. As CA = OM = r and OP = 2, PM = r-2.
Triangle ∆CMP:
CM² + PM² = CP²
(196-28r) + (r-2)² = r²
196 - 28r + r² - 4r + 4 = r²
32r = 200
r = 200/32 = 25/4
Circle C:
Aᴄ = πr² = π(25/4)² = 625π/16 ≈ 122.72 sq units
Excellent!
Thanks for sharing ❤️
Essa questão foi maravilhosa!
Gracias, querido❤️
Sorry, no idea: Why are O, C and B collinear, particulary O ??
They are both tangent at point B. So by def, the line from their center to B is perpendicular (90 angle) to the tangent line. If they both have a perpendicular line at B going to their center, then the lines are colinear, and OCB is a straight line.
@@MegaSuperEnrique Thank you very much, I had an obvious mistake in thinking
r²=x²+(r-2)²
Thanks for the feedback ❤️
All Hail Jakob Steiner! ...and contribution of Point of Infinity. 🙂
😀❤️
Thanks for the feedback ❤️
14 is the radius of the quarter circle, ....it is difficult puzzle😅
Thanks for the feedback ❤️