The point-normal form of equation of a line (i.e., n·x = n·p) seems more convenient and less error-prone in this application. For instance, to find the equation of the perpendicular bisector of AB, we compute n = B-A = (7,-1) and the midpoint p = (A+B)/2 = (3/2,3/2), yielding (7,-1)·(x,y)=(7,-1)·(3/2,3/2) or 7x-y=9.
Alright I'm feeling charitable so if any of you noobs are finding this hard then may I suggest a more formulaic approach? 1) Consider the general equation for a circle : *(x−x₁)²* *+* *(y−y₁)²* *=* *r²* Plug in the three points to create three quadratic equations. 2) Now there should be 3 equations like: *(x−a)²* *+* *(y−b)²* *=* *r²* *①* *(x−c)²* *+* *(y−d)²* *=* *r²* *②* *(x−e)²* *+* *(y−f)²* *=* *r²* *③* 3) Subtract the *①* from the *②* which will create another equation that's linear, call it ④ and subtract ① from ③, call it ⑤ 4) Solve the two Simultaneous linear equations: *ax* *+* *by* *=* *c* *④* *dx* *+* *ey* *=* *f* *⑤* which will give you the center coordinate points of the circle. 5) Last step is to plug the center points into any of the ①, ② or ③ equations and solve for *r* and the rest is history.
The 3 quadratic equations forms 3 circles that all intersect at the center of the main circle which you're trying to find, after you derive 2 straight line equations from the non-linear circle equations those lines also intersect at the center of the circle which ultimately provides you the center coordinates of the circle.
@@michellesilva7366 all the three equations are of the same circle, so obviously the radius in each equation is the same and is that of the circle that we are trying to find!
Yeah. You can use this: gradient of eqn (m1) times perpendicular gradient (m2) = -1. If you make the perpendicular gradient the subject, then you can see how it changes signs and 'flips'.
1 year late so this is mainly for anyone else reading but you can use the method in the vid for a line from the midpoint of AB through the centre and the question should give the equation of the tangent/ 2 points on the tangent, so you can do the negative reciprocal at A then set the two equal and solve.
Are you referring to the chords? In this example, the chord AB is not horizontal., and its perpendicular bisector is not vertical. The only formula required for this question is the equation of a circle (x-a)^2 + (y-b)^2 = r^2, where (a,b)is the centre of the circle and r is the radius. The centre (a,b) is found at the intersection between the perp bisectors of chords AB and AC.
This is a standard method of finding the standard equation of the circle passing 3 points as opposed to the standard equation of the circle passing through 3 points.
@@zachpaterson8128 It is not a coincidence. It depends on what you call X2 and Y2. Somebody else may call your X2, Y2 there X1, Y1. If you multiply top and bottom of the fraction by -1 you will see that (Y2-Y1)/(X2-X1) = (Y1-Y2)/(X1-X2)
![A-Level Maths: C2-11 [Circles: Given 3 Points, Find the Centre & Radius - Example 1]](/img/1.gif)
Why do I love exam solutions? Simply it’s amazing especially when you’re stuck in a question and have no one to ask!!!!!
you sound happy as mate u really make revising fun . u the man appreciate ya !
Outstanding explanation! The intersection of the chords' perpendicular bisectors is the key!
i think ill actually get this question right on exams now thank you
The point-normal form of equation of a line (i.e., n·x = n·p) seems more convenient and less error-prone in this application. For instance, to find the equation of the perpendicular bisector of AB, we compute n = B-A = (7,-1) and the midpoint p = (A+B)/2 = (3/2,3/2), yielding (7,-1)·(x,y)=(7,-1)·(3/2,3/2) or 7x-y=9.
ahhh such an amazing video thank you!!
Glad you liked it!!
Could you please go through the edexcel a level 2019 mock exams please
Thank youu from Bangladesh
you are the best you have made me understand this concept
You videos are very helpful sir. I hope you continue to upload more content. From India!
Thank you. Yes I hope to continue providing people keep supporting this work.
Thank you for making maths videos for us! You have helped me raise my grades a lot! ❤️
Thank you so much. This helped me a lot. :)
That's good. Best wishes
excellent explanation my brother.....
Outstanding; thank you
Alright I'm feeling charitable so if any of you noobs are finding this hard then may I suggest a more formulaic approach?
1) Consider the general equation for a circle : *(x−x₁)²* *+* *(y−y₁)²* *=* *r²*
Plug in the three points to create three quadratic equations.
2) Now there should be 3 equations like:
*(x−a)²* *+* *(y−b)²* *=* *r²* *①*
*(x−c)²* *+* *(y−d)²* *=* *r²* *②*
*(x−e)²* *+* *(y−f)²* *=* *r²* *③*
3) Subtract the *①* from the *②* which will create another equation that's linear, call it ④ and subtract ① from ③, call it ⑤
4) Solve the two Simultaneous linear equations:
*ax* *+* *by* *=* *c* *④*
*dx* *+* *ey* *=* *f* *⑤*
which will give you the center coordinate points of the circle.
5) Last step is to plug the center points into any of the ①, ② or ③ equations and solve for *r* and the rest is history.
The 3 quadratic equations forms 3 circles that all intersect at the center of the main circle which you're trying to find, after you derive 2 straight line equations from the non-linear circle equations those lines also intersect at the center of the circle which ultimately provides you the center coordinates of the circle.
Woooooowww thanks a lot!
But how do you find the radius of all 3 equations?
Hey bruh I need your help
@@michellesilva7366 all the three equations are of the same circle, so obviously the radius in each equation is the same and is that of the circle that we are trying to find!
Thans mate
Thanks for saving my life, sir
Good to hear. Best wishes.
Hey! Can I ask about the gradients? If the gradient is positive, does it become negative as a perpendicular gradient?
Yeah. You can use this: gradient of eqn (m1) times perpendicular gradient (m2) = -1.
If you make the perpendicular gradient the subject, then you can see how it changes signs and 'flips'.
Thanks for the help❤️❤️❤️❤️!!! I have my As maths exam tomorrow
Thank you
Chrystal-clear explanations; wonderful!
Is there a way to calculate a circle from point A, point B and a tangent passing through point A?
1 year late so this is mainly for anyone else reading but you can use the method in the vid for a line from the midpoint of AB through the centre and the question should give the equation of the tangent/ 2 points on the tangent, so you can do the negative reciprocal at A then set the two equal and solve.
How many marks would these sort of questions be worth in a typical exam ?
nice work
Can you please do worked solutions for M1 IAL past papers? Please, I really need your help. All of your C12 solutions were super useful
This help a lot !! Thanks ;)
Excellent concept . Keep going, very helpful
What if it doesnt have vertical or horizontal line? What formula is needed?
Are you referring to the chords? In this example, the chord AB is not horizontal., and its perpendicular bisector is not vertical. The only formula required for this question is the equation of a circle (x-a)^2 + (y-b)^2 = r^2, where (a,b)is the centre of the circle and r is the radius. The centre (a,b) is found at the intersection between the perp bisectors of chords AB and AC.
@@ExamSolutions_Maths oh so like we could still do the same thing even if the chord is not vertical?
😊 thank you so much
Thank you sir super teaching
HAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAHHHH, thanks been dying over a sum like this
Can i apply this same logic with 4 points given?
Is this still relevant to the AQA spec? I’ve not come across it at all
i do edexcel, but i have only come across a question with 2 points
You're a lifesaver ❤️
Thank you very much this help me
Much better update of this question in higher resolution
FINALLY someone who's not American 🎉
Thank you soo much
2:57 how did you know that was the center?
The centre of a circle always lies at the midpoint of any perpendicular bisector of a chord of a circle.
I just want to ask if this is the standard equation of the circle passing 3 points
This is a standard method of finding the standard equation of the circle passing 3 points as opposed to the standard equation of the circle passing through 3 points.
@@ExamSolutions_Maths thank you 😘😘
Nailed it in 5:30 (5 mins 30 seconds)
Yo thanks dawg ya help me
how many marks do you think this question would be worth in an exam?
I might be going crazy but didn't you do the slope formula backwards? Normally it's (Y2-Y1)/(X2-X1)
You and I are correct. It will give the same answer, try it.
@@ExamSolutions_Maths Does it always work both ways? Or is this just one of those happy coincidences?
@@zachpaterson8128 It is not a coincidence. It depends on what you call X2 and Y2. Somebody else may call your X2, Y2 there X1, Y1. If you multiply top and bottom of the fraction by -1 you will see that (Y2-Y1)/(X2-X1) = (Y1-Y2)/(X1-X2)
@@ExamSolutions_Maths their*
Thanks my name is murad and my teacher has an impala the size of my future
A what?!?
Sir
STEM PA MORE
this is too complicated
I think u made the gradient ab wrong its y_2-y_1 not y_1-y_2
It comes out the same answer.