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Circle

Locus of points at a fixed distance, r, from point P is known as a circle. In this case, distance r is known as radius and point P is centre.
Description: 15554.png

Equation of a Circle

If the coordinate of the centre is (ab) and length of radius = r, then equation of circle is:
 
(x - a)2 + (y - b)2 = r2
 

Though, equation of circle may not be always given in the above format. Consider the following equation:
 
x2 + y2 - 2x - 4y - 4 = 0
 
It should be written as a summation of (x - a)2 and (y b)2
 
First of all, collect terms of x in one bracket and terms of y in other bracket.
 
(x2 - 2x) + (y2 - 4y) - 4 = 0
 
Then start writing the terms in the bracket as squares, keeping the constant terms adjusted.
(x2 - 2x + 1) + (y2 - 4y + 4) - 4 - 5 = 0
⇒ (x - 1)2 + (y - 2)2 = 32

Hence, this circle has its centre at (1, 2) and its radius = 3 units.

 

General Equation of the Circle
  1. The general equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0
     
    Centre of this circle is (-g, -f ) and its radius = Description: 14620.png
  2. If centre of the circle is Origin (0, 0), then equation of circle is: x2 + y2 = r2
 
Example-1
Find the radius and centre of the circle x2 + y2 - 6x + 4y - 12 = 0.
Solution
Centre of the circle x2 + y2 + 2gx + 2fy + c = 0 is = (-g, -f)
 
Comparing this with the equation given in the question:
Description: 15578.png
 
2g = -6 ⇒ -g = 3
 
2f = 4 ⇒ -f = -2
 
Hence, centre = (-g, -f) = (3, -2)
 
Radius = Description: 14627.png = Description: 14634.png = 5 units.
 
 
Example-2
Find the radius and centre of the circle 2x2 + 2y2 - 8x - 7y = 0.
Solution
We are required to first write it down in the format such that co-efficient of each of x2 and y2 = 1.
 
2x2 + 2y2 - 8x - 7y = 0 can be written as x2 + y2 - 4x - Description: 14644.png y = 0.
 
In this case, g = -2 and f = Description: 14655.png
 
Hence, centre = (-g, -f) = Description: 15595.png
 
Radius = Description: 14671.png = Description: 14679.png = Description: 14692.png
 




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