Finding centre of circle on an Argand diagram, given 3 points

[H.Bisho18]

Hi,

So I've done all the work correct up to d which i am stuck on.

the 3 points are. Origin, 3+2i, 2-3i. I was thinking i have to do something with 'mod' z-(a+bi) 'mod' = radius but then part e is find out the radius so now i'm confused.
Thanks for any help given

 

[pka]

the 3 points are. Origin, 3+2i, 2-3i. I was thinking i have to do something with 'mod' z-(a+bi) 'mod' = radius but then part e is find out the radius so now i'm confused.

Any three non-colinear points determine a circle. Here you have \(\displaystyle (0,0),~(3,2)~\&~(2,-3)\)
Look at this link. You should learn to use that resource.

 

[H.Bisho18]

Any three non-colinear points determine a circle. Here you have \(\displaystyle (0,0),~(3,2)~\&~(2,-3)\)
Look at this link. You should learn to use that resource.

Cool, however without using that resource how would i find the centre?

 

[lev888]

Circle equation: (x – h)² + (y – k)² = r²
You need to find h, k, r. Luckily, you have 3 points you can use to make a system of 3 equations.

 

[pka]

cool, however without using that resource how would i find the centre?

Well, it is tedious in the extreme.
Let \(\displaystyle \mathit{O} : (0,0),~\mathit{A} : (3,2)~\&~\mathit{B} : (2,-3)\)
Write the perpendicular bisector of \(\displaystyle \overline{\mathit{O}\mathit{A}}\) and the perpendicular bisector of \(\displaystyle \overline{\mathit{O}\mathit{B}}\).
Those two lines intersect at the centre of the circle call it \(\displaystyle \mathrm{C}\)
Now the length of \(\displaystyle \overline{\mathrm{C}\mathit{A}}\) is the radius.

 

[Dr.Peterson]

This is much easier to solve geometrically. They told you to draw the diagram; that will at least suggest (and you can then demonstrate) that this is a right isosceles triangle (part of this is what you were told to prove in part c). You may know something about the circumcircle of any right triangle (that is, where the center of a circle is, that passes through the vertices); if not, you could add in a fourth point to form a square, and think about where its center is.

This is a good example of the fact that algebra is not always the best way to solve a problem, and that taking the time to think about a problem from multiple perspectives can help in finding the least tedious way.

In this case, just doing exactly what they say leads to this easy method! Textbook authors are not always trying to make things hard for you ...