Find the Radius of a Circle when Given 2 Perpendicular Chords | Quick & Simple Tutorial
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- Опубліковано 29 сер 2024
- Learn how to find the radius of a circle when only given the lengths of two perpendicular chords. Utilize the chords theorem, chord of a circle theorem, and the pythagorean theorem. Step-by-step explanation by PreMath.com
I love these videos. I built a different right triangle with the radius hypotenuse.
Once you determine that AP=7, you know that PF and CN=1; and that NE=8.
CE^2=8^2+1^2=65
CE=8.06
How clever such solution is! I was astonished by the solution uploaded and was amazed all along while watching it. I will watch it again and other videos of yours. So nice, so clear.
This was a nice question and a very easy to follow explanation about getting to the right answer. At the 5:29 time point of the video, an alternative path opens up to get the result:
(CD)^2 = (CN)^2 + (DN)^2
and substituting the values (because CD is the radius) we get to the same answer:
r^2 = 1^2 + 8^2 = 65.
And the same result is obtained if the triangle CNE is used instead.
I got this one using the formula 4r^2 = 8^2 + 12^2 + 6^2 + 4^2. Got 8.06. AF = 8 using intersecting chord theorem. Let's keep rolling bro, love these brain teasers.
Thanks my friend!
same method i did
YES RIGHT
Can you please explain me the formula .
I like to say it is not for all types of intersecting chords.
Only when they intersect perpendicularly only, then this formula is feasible.
I have been teaching Maths for a very long time. I love many of your videos because it makes me challenge areas of Geometry I thought I knew. On this problem, it suddenly became apparent that where you have two intersecting perpendicular chords, the circle centre must be at right angles to the midpoint of each chord. Never realised that before. I think that the teaching of good old fashioned Geometry is something that is often weak in the classroom.
Straight forward way is choose point E as =0,0.
Point B=6,4
Point D=0,16
Use equation or a circle 3 times
Where Xc and Yc are coordinates of center of circle.
(X-Xc)^2+(Y-Yc)^2=R^2
First point E
(0-Xc)^2+(0-Yc)^2=R^2
Point B
(6-Xc)^2+(4-Yc)^2=R^2
Point D
(0-Xc)^2+(16-Yc)^2=R^2
Then use algebra from there on.
I used intersecting chord theorem again after finding cn=1.
ED is bisected by diameter at 90 degrees.
Diameter is bisected into lengths r+1 and r-1.
(r+1)(r-1)=ENxDN=8x8
r^2 -1=64
r = sqrt 65
Love the videos. Learning a lot
Nicely done! Could also use the triangle NDC, fortunately 4^2 + 7^2 == 1^2 + 8^2 so you DO get the same answer! :-))
Using coordinate geometry can simplify things a little, although exactly the same calculations are made. Calculate AF = 8 as before using intersecting chords theorem.
Set the origin at A and the X-axis along AB. Again, use the fact that the perpendicular bisector of a chord passes through the centre and consider chords AB and DE.
The X-coordinate of C is therefore half the X-coordinate of B. That gives 14/2 = 7.
The Y-coordinate of C is the mean of the Y-coordinates of D and E. That gives (12 + (-4))/2 = 8/2 = 4.
The distance from the origin (A) to a the point (7, 4), which is C, is given by sqrt(7^2 + 4^2) = sqrt(49 + 16) = sqrt(65). That is the radius.
As you can see, it requires less drawing, but still requires the same calculations. Some students will take to the geometrical approach better, while others may find the coordinate method more to their liking. There's no "wrong" method here.
I got it by using formula 4r²=12²+8²+6²+4²
r= root 65
Got AF=8 by chords' theorem formula.
You post very good questions, and provide very nice solutions as well. ❤️
Comments like this are exactly why I read them. I found the 8 & made a 2 x 16 box, then use pythagoras, but I really like this new to me formula.
Thanks for posting.
Used the same way. One should point out though, that the formular 4r² = w²+x²+y²+z² only applied when the chords are perpendicular to oneanother. This Video ua-cam.com/video/cSkswi-Y0x4/v-deo.html explains how to get to this formula.
You could also use the formula for the cord length:
cord length = 2 × √(r²− d²) with r= radius circle and d=distance from the centre of circle perpendicular to the cord (here CN or CP).
We can rewrite this formula as:
(cord length)² = 4(r² − d²) = 4r² - 4d² and so 4r² = (cord length)² + 4d² which gives r² = 1/4 x (cord length)² + d².
So if we take the cord DE:
r² = 1/4 x DE² + d² = 1/4 x 16² + d² = 64 + d².
d = CN = AF - (AB/2) = 8 - 7 = 1 so we get r² = 64 + d² = 64 + 1 = 65 which gives us r = √65.
If we take cord AB we'd get:
r² = 1/4 x 14² + d² = 49 + d². Here d = CP = DF - (DE/2) = 12 - 8 = 4 and so r² = 49 + 16 = 65 which gives us r = √65.
Thank you for a fantastic solution!
My aproach is nearly the same like yours at first, but it is a bit different.
First I calulate the lenght of CF: sq CF = sq1+ sq 4+ 1+16= 17---> CF= sqrt of 17.
Extend CF to the other sides of the circle to have a diameter (radius=R), using the chord theorem, we have: (R+CF) x(R-CF) = FExFD
----> sq R- sq (sqrt of 17) = 4x12 = 48----> sq R -17=48 ----> R= sqrt of 65 units
Using coordinate geometry let's put the center in F and calculate the circle passing for 3 points D(0,12) B(6,0) and E(0,-4), knowing that r = sqrt[(-a/2)^2+(-b/2)^2-c]
144 + 12b + c = 0
36 + 6a + c = 0
16 - 4b + c = 0
that gives a=2, b=-8 and c=-48, then r=sqrt65
OK, so you find that
AF = 8; AP = 7; 👍
⇒ FP = 1
And then DN = 8
But FP = r - x = 1, where x is the rest of the radius along the CN direction.
⇒ x = r - 1
So, why don't we apply the intersecting chords theorem once more to get:
(r + 1)·(r - 1) = DN·NE = 8²
(for DE and its perpendicular segment along CN)
⇒ r² - 1 = 8²
⇒ r² = 64 + 1
⇒ r = √6̅5̅
I used this method as well
We can also use this formula as if we know length of segment AF by the formula intersecting chord theorem: (AF)(FB)=(DF)(FE), and we get AF=8. After getting this one we then use formula for finding the radius(r), that is 4r²=AF²+FB²+DF²+FE², now we fill up the blanks we get: r²=(8²+6²+12²+4²)/4
r²=260/4
r²=65
r=√65
r=8.06225775.
🙏❤️
Excellent Don
You are awesome 👍
Take care dear and stay blessed😃 Love and prayers from the USA!
@@PreMath You're most welcome from India🇮🇳🇺🇸
Thanks a lot Sir
🙏❤️
That is terrific, but where did this formula come from? Does it have a generic name?\
@@theosalt1 4r^2 is the area of big square with 2r as length of the side of the square.. Then we also can calculate 12^2 + 6^2 + 4^2 + 8^2 as the area of the big square..
So that 4r^2 = 12^2 + 6^2 + 4^2 + 8^2.
Then r = (square root)65.
CMIIW
I am so sorry, but I need to say that
you are wrong, r=√65, but r≠8.06225775, since the latter is just approximate value.
in Chinese saying, your behavior is called "畫蛇添足". It is a pity.
Segment CN is 1, segment NE is 8. No need for complications, you have Pythagoras satisfied then.
let F(0,0), B(6,0), D(0,12), E(0,-4)
the slope of chord BD is (-12 / 6)=-2, and the midpoint of BD is ((6+0)/2 , (0+12)/2)=(3,6)
so the perpendicular bisector of BD is (y-6) / (x-3)=1/2, namely x-2y=-9
the slope of chord BE is 4/6=2/3, and the midpoint of BE is ((6+0)/2 , (0+(-4))/2)=(3,-2)
so the perpendicular bisector of BE is (y-(-2)) / (x-3)=-3/2, namely 3x+2y=5
the two perpendicular bisectors intersect on the point (-1 , 4), namely C locates on (-1 , 4)
the radius is CD=√((-1-0)^2+(4-12)^2)=√65
done!!
Other way (to show the advantage of analytic geometry)
M middle DE
N middle BE
O circle centre ;
r = CB = CE
if M(0,0) :
F(0,-4) ; B(6,-4) ; E(0,-8) ; N(3,-6)
find centre C(x,0)
1) on graph
perp.bis. of chord BE (by middle N)
cuts x-axis CM (perp.bis. of chord DE) in C(-1,0)
2) arithmetic
C(x,0) ; B(6,-4) ; E(0,-8)
CB = CE
=> (6-x)² + (-4-0)² = (0-x)² + (-8-0)²
=> -12 x = 12
=> x = -1 => C(-1,0)
r = CB (= CE)
r² = (6+1)² + (-4-0)²
r² = 65
r = √65
A solution by Coordinate Geometry
Point F=(0 , 0)
Join AC, BC & DC (all are radii).
A=(-8,0) B= (6,0) D =(0,12)
Let C= (x , y)
AC^2 =(x +8)^2 + y^2 --(1) BC^2=(x -6)^2+y^2--(2)
CA^2 =x ^2 +(y-12)^2 -(3)
From equations (1) & (2)
(x+8)^2=(x-6)^2
x=-1
From (1) & (3)
49 +y^2 = 1+(y -12)^2
>24y = 144 - 48=96
y =4
(x, y ) =( -1 , 4)
OB =√[(6+1)^2+ (0 -4)^2]
=√65
Radius =√65 unit
Comment please.
Got the correct answer using cosine rule. Your method is much cleaner.
It is easier to calculate the radius from the triangle "CND" sİnce you know CN=1 and DN=8..
Using theory that angle at centre is double the angle at circumference of circle, we can then use similar triangles of half the triangle of CEB and triangle DFB.
We have EB = sqrt 52, so half of that is Sqrt 52/2.
Also DB = sqrt 180,
So:
Sqrt 52/2r = 6/Sqrt 180
r= Sqrt 65.
Short and simple
The solution was so obvious I got it after 10 seconds looking at the problem without writing down anything.
intersecting chords: 6x=12*4: x=8;
0.5(8+6)=7 midpoint
0.5(12+4)=8 midpoint
point c at 7, 4. let that be a rt triangle
with hypotenuse chord ac
49 + 16 = hyp squared
65 = hyp squared
8.06= hyp
you did it exactly the same way.
Anybody else use the theorem of angles in a semicircle are 90 so copy the bottom chord and move it up in the circle so in effect you are creating a box 14x8. The diagonal of this box must by the diameter since there is a right angle (angles in a semi circle are 90) so just solve using (2r)^2 = 8^2+(8+6)^2 r=sqrt65
Hi dear.
We can solve this problem much simple. Just draw a parallel line over/above AB equal in length to
E.g. MN ll AB. Thus we have got a rectangle MNAB - inscribed to the circle.
Now we can simply consider a triangle, e.g. ABN. The hypotenuse AN is gonna be the diameter of the circle. So, we have two known cathetus AB and BN and potegrian theorem diameter=√(AB^2 + BN^2) .
From where diameter turn out to be √260 or 2√65.
Here we are 👍
CP perpendicular to AB and
CN perpendicular to DE be drawn.
So AP = BP
= (BF + EF*DF/ BF)/2
= (6+12*4/6l/2 = 7
So CN = 8-7 = 1
and radius ^2 = CN^2 + NE^2
= 1 + 8^2 = 65
Yes, no need to find out CP.
Another great example 👍🏻
I drew the radius line from C to D. Fun problem. Thanks.
Thank you sir. Your explanation is great. I love it
Woo hoo! I got one exactly as shown before the explanation!
Me too
The interesting thing is that if radius is 8, and the chord ED is 16, than the chord ED is a diameter.
Just go 4 down from D and then 8 to the left to the edge of the circle and call this point H.
The diameter HB is then sqrt((8+6)^2 + 8^2) = sqrt(260) = 2*sqrt(65).
Beautifully done Sir 😇👏👏👏
You could have done the same thing -- same concept -- by figuring out the hypotenuse of right triangle, CDN where CD would be the hypotenuse.
This question can also be solved using coordinate geometry using general equation of circles by using three coordinates three equations will be made and radius of circle can be found
The problem is solved very simply by using the sine theorem for the AED triangle: R=AE/(2*SinD).
That's super amazing ... ❤❤ and logical
So nice of you Sangam Thank you so much for taking the time to leave this comment.! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃
Nice. I used coordinate geometry to solve it.
Boa aula!
From the figure it is clear the coordinates of the three points of the circle (0, 12) (6, 0) ( 0, -4).
What is the problem?
4r^2 = 8^2 + 12^2 + 6^2 + 4^2 .... r=8.06. Center - (-1, 4)
Please make some questions based on Stewart theorem, Ptolemy theorem,area of triangle using area concepts and on cyclic quadrilateral
Wow super explanation👍
Thank you so much for your hard work sir😀
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Good... Simple and enjoyable.. I like it
If ab=cd wouldn't ab and de be congruent? Meaning that af=10?
I was kind of astonished to find that it's a requirement for the chords to be perpendicular. I would like any reference to a mre general case (dissected chords arranged any way.
Thank you
Why do you mark the two parts of DE as if they were equal? I was taught that.
I solved by general equation of circle, a longer way, anything, I like your video, detailed and good explained, greetings!
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4^2 + 7^2 = c^2
16 + 49 = 65
c^2 = 65
c = 8.06 (about 8.1)
Awesome, many thanks! Take the other side you get 8^1 + 1 = 65 → r = √65 🙂
Or take the fast lane: 4r^2 = (AF)^2 + (BF)^2 + (DF)^2 + (EF)^2 = 64 + 36 + 144 + 16 = 260 →
r^2 = 65 → r = √65 🙂
Can drop down radius cd on side cn,dn=8,cn=1 and apply theorem paphahorean theorem,I am from Russia,sorry pre my English
Great Video
Very good
Thanks JSasmal dear for the visit! I
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
I used coordinate plane to solve it I got the same answer
12squared+4squared+6squared+8squared=4times radius squared. i.e. 260=4times r squared i.e. r=square of 65. It's just using a formula.
Wao ,I subscribed ur channel
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4R२=A२+B२+C२+D२
where A=4, B=12, C=6, D=8
Radius = 4_/39 divided by 3 (ans)
Thank you sir
So nice of you dear!
By which theorem
👍
r=√65≈8,06
How ab=cd?
解けた！
すごい！とても賢い👍大切に気をつけて、祝福されてください😃
R^2=20
Should have written r= square root of 65 above
bye ! 😄
🤔👍
Right triangle APC
#Radius
#RightTriangle
#RightTriangle #Pythagoras #PythagoreanTheorem
Right triangle
R = sq.rt 65
Thank me later:
DF x FE = AF x FB
4; 7; square root of 65
#Pythagoras #PythagoreanTheorem
Pythagorean Theorem
Bài toán cũ
r = Square root of 65
Radius = Square root of 65
Square root of 65
Pythagoras
R=8.062257748
Please translet to Spanish.....
Thanks Jorge for suggestion!
So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
4; 7; square root of 65
Pythagorean Theorem
Radius = Square root of 65
Pythagoras