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The main question is as follows :

Find the point $P$, such that $P$ is the centre of a circle circumscribing the triangle whose angular points are $(1,1), (2,3), (-2,2)$.

My method :

I am completely new to Coordinate geometry of higher level. I understood what the question asks, it asks me to find the circumcenter of the triangle, but all the formulae given to me are very confusing. I'm having trouble proceeding further. Please help me.

Yuriy S
  • 31,474
  • If you are completely new to analytic geometry, then you wouldn't be expected to solve this problem. If you are expected to solve it, then you were given the necessary tools. Please, provide the "formulae given" to you, at least, and tell what is unclear to you about them – Yuriy S Jul 21 '16 at 12:41
  • By the way, what did you mean by "confusing language"? – Yuriy S Jul 21 '16 at 12:45
  • If the given points are $A,B,C$, the Center $P$ is the point of intersection of the perpendicular bisectors of $AB$ and $BC$. –  Jul 21 '16 at 13:22

1 Answers1

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The general equation of a cirlce with the center $x_0,y_0$ and radius $R$ is:

$$(x-x_0)^2+(y-y_0)^2=R^2$$

You know three points which lie on the circle. Thus, you can create a system of three equations to find the three unknowns $x_0,y_0,R$:

$$ \begin{cases} (1-x_0)^2+(1-y_0)^2=R^2 \\ (2-x_0)^2+(3-y_0)^2=R^2 \\ (-2-x_0)^2+(2-y_0)^2=R^2 \end{cases}$$

Yuriy S
  • 31,474
  • Is there any shorter method to do this? –  Jul 21 '16 at 13:26
  • @AksharGandhi, why do you need a shorter method? This one is extremely short and easy. Just expand the squares, and subtract (for example) the first equation from the second, and the second equation from the third. You will get a system of two linear equations for $x_0,y_0$ – Yuriy S Jul 21 '16 at 13:50
  • Yes it is. Thanks a lot @You'reInMyEye! –  Jul 21 '16 at 15:24