SOLUTION: Find a point (x,y) equidistant from points (2,-7), (-3,-8) and (-5,9)

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Question 1128770: Find a point (x,y) equidistant from points (2,-7), (-3,-8) and (-5,9)
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
Let the points be  A = (2,-7), B = (-3,-8) and C = (-5,9).



    1.  Write an equation of the straight line which is a perpendicular bisector to the segment AB.


    2.  Write an equation of the straight line which is a perpendicular bisector to the segment BC.


    3.  Find the intersection point of these two straigh lines by solving the system of these two equations.


    4.  This intersection point is your answer.

        It is the center of the circle circumscribed around these three points.

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If you need more explanations or if you want to have a TEMPLATE before your eyes, look into the lesson
    - Find the standard equation of a circle
in this site.


Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find a point (x,y) equidistant from points (2,-7), (-3,-8) and (-5,9)
==========
Find the equation of the circle thru the 3 points.
The center of the circle is equidistant from the 3 points.
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87x^2 + 373x + 87y^2 - 125y = 6232 is the circle.
Complete the squares for x and y to find the center.

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