Example 2 : Equation of a Circle given 3 points on the Circumference : ExamSolutions

[tomwann1]

thank you this helped me alot, appreciate this keep up the good work

[64adamw64]

How am I meant to find the centre of a circle given only 2 points on the circle (a chord) and a tangent??

[ExamSolutions]

@64adamw64 Find out where the normal to the tangent and the perpendicular bisector of the chord intersect. You will need to establish the respective equations and solve simultaneously.

[64adamw64]

@ExamSolutions ok thanks :)

[tehemor]

Will the point of intersection of the perpendicular bisectors always be the centre point or is it just this specific example?

I'm pretty sure it is 'cause when I visualise it in my head it seems to be so but I don't know for sure...

Thanks for the videos, they're very helpful.

EMZ=]

[ExamSolutions]

@tehemor Yes

[MilitaryMan006]

another question, if you dont have a vertical chord can you still find the midpoint of the chord and perpendicular gradient and work it out via simultaneous equation?

[MilitaryMan006]

how can you do a neat construction of the circle during the exam without knowing the centre?

[ExamSolutions]

@MilitaryMan006 practice and guess work

[Goodfellow7]

When I first read the problem, I quickly drew a small scale diagram and clearly saw that the centre was (1, -2) before I even started doing anything. Would I lose marks in an exam if I just solved the problem from there?

[ExamSolutions]

@Goodfellow7 I would think so. I would encourage you to work it out algebraically. Suppose the real answer was (1.1, -2.1) or very close to your answer. Do you think you would have been able to spot this?

[bizayangdako]

can u also include the general equation?

[sblueninja]

Wait.. when you solved the radius, it came out as this: r² = 25. Isn't it supposed to be r=√25 ? Just curious. :)

[ExamSolutions]

@sblueninja Nope, I didn't solve it for r but r^2. r^2=25 so r=5 but the equation ends in r^2 so it should be 25.

[ijam1234567890]

thanks:)

[ExamSolutions]

@ijam1234567890 pleasure

[ijam1234567890]

what if only two points are given and they are not vertically above each other?

[ExamSolutions]

@ijam1234567890 You must always have 3 points. If one is not above the other or horizontal to the other then the answer is longer. Find the equations of the two perpendicular bisectors and see where they intersect by using simyultaneous equations. The point is though, always sketch a diagram so that any obvious points can be seen. It may save you a lot of work.

[JustCut08]

Do you realize that your nothing short of amazing?

You have help me understand many things, please keep it up. You are educating people in such a good way. You should be proud :)

[ExamSolutions]

Thanks - Comments like yours always give me encouragement.

[amyyanwatse]

Thank you very much for the making of the videos! I find them extremely helpful :D

[ExamSolutions]

That's good and thanks for the donation

[briancarrane]

Helped me alot