Tuesday, July 15, 2014

Analytical Geometry : Circle Equations

Circle Equations from Math is Fun

How to Find Equation of a Circle Passing 3 Given Points

7 methods included ; Amazing !!

Practice finding the equation of a Circle given 3 points --

Q #1 : (1, 3), (7, 3) and (1, -3)

Answer : (x -4)2 + y2 = 18
Q #2  : (3, 4), (3, -4), (0, 5)
Answer : x2 + y2 = 25
Q #3 : A (1, 1), B (2, 4), C (5, 3)
Answer : (x-3)2 + (y -2)2 = 5
Solution :
The midpoint of line AB on the Cartesian plane is $$(\frac{3}{2}, \frac{5}{2})$$ and the slope is $$(\frac{3}{1})$$ so the slope of the perpendicular bisector of line AB is $$(\frac{-1}{3})$$.
The equation of the line bisect line AB and perpendicular to line AB is thus :
y - $$(\frac{5}{2}$$) =$$\frac{-1}{3}$$ [x - $$(\frac{3}{2})$$] --- equation 1
The midpoint of line BC on the Cartesian plane is $$(\frac{7}{2}, \frac{7}{2})$$ and the slope is $$(\frac{-1}{3})$$ so the slope of the perpendicular bisector of line BC is 3.
And the equation of the line bisect line BC and perpendicular to line BC is
y - $$(\frac{7}{2})$$ = 3 [x - $$(\frac{7}{2})$$] --- equation 2
Solve the two equations for x and y and you have the center of the circle being (3, 2)
Use distance formula from the center circle to any point to get the radius =
$$\sqrt{5}$$.
The answer is : (x - 3)2 + (x - 2)2 = 5

More practices on similar questions :  (Answers below for self check)

Q #1 : A (2, 5) , B (2, 13) ,  and C (-6, 5 )

Q #2 : A (0, 7), B ( 6, 5 ), and C (-6, -11 )

Q #3 : A (3, -5) , B (-4, 2) and C (1, 7 )