To find equation of circle which passes through three different points, we have different ways.
Here we see how to find equation of circle from perpendicular bisectors.
Step 1 :
By connecting any of the two points on the circle, we will get chord of the circle.
Step 2 :
For the chord AB, find midpoint and slope of the chord AB. To find equation of perpendicular bisector of the chord AB, we may use this point derived from the points A and B and finding the negative reciprocal of the chord.
Step 3 :
The same steps should be done for another chord also.
Step 4 :
By finding the point of intersection of these two chords, we will get the center of the circle.
Step 5 :
We can find the equation of circle using the formula
(x - h)^{2} + (y - k)^{2} = r^{2}
Problem 1 :
Three points on the circle:
(-18, -5), (-7, -16) and (4, -5)
Solution :
The perpendicular bisector of two chords will meet at center.
Let A (-18, -5) B (-7, -16) and C (4, -5) be the points on the circle.
Midpoint of AB :
A (-18, -5) B (-7, -16)
Slope of AB :
Slope of perpendicular bisector = 1
Midpoint of BC :
B (-7, -16) and C (4, -5)
Midpoint = (-7 + 4)/2, (-16 + (-5))/2
= -3/2, -21/2
Slope of BC :
Slope of perpendicular bisector BC = -1
Point of intersection of two perpendicular bisector AB and BC,
x - y = -2 ---(1)
-x - y = 12 -----(2)
(1) + (2)
-2y = 10
y = -5
Applying the value of x in (1), we get
x + 5 = -2
x = -7
Center is at (-7, -5)
Distance between center (-7,-5) and a point on the circle (-18, -5)
Radius = √(-7 + 18)^{2} + (-5 + 5)^{2}
= √11^{2} + 0^{2}
= √121
r^{2} = 121
(x + 7)^{2} + (y + 5)^{2} = 121
x^{2} + 14x + 49 + y^{2} + 10y + 25 = 121
x^{2} + y^{2} + 14x + 10y + 49 + 25 - 121 = 0
x^{2} + y^{2} + 14x + 10y - 47 = 0
So, the equation of circle is x^{2} + y^{2} + 14x + 10y - 47 = 0.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM