Equation of a circle in standard form with center (0, 0) and radius 'r'.
x2 + y2 = r2
Equation of a circle in standard form with center (h, k) and radius 'r'.
(x - h)2 + (y - k)2 = r2
Equation of a circle in general form :
x2 + y2 + 2gx + 2fy + c = 0
where the center is (-g, if) and radius is √(g2 + f2 - c).
Find the center and radius of the following circles.
Example 1 :
x2 + y2 = 25
Solution :
The given equation of the circle is in the form of
x2 + y2 = r2
So, the center of the given circle is (0, 0).
Radius :.
r2 = 25
r = 5 units
Example 2 :
(x - 1)2 + (y - 3)2 = 9
Solution :
The given equation of the circle is in the form of
(x - h)2 + (y - k)2 = r2
Center :
(h, k) = (0, 0)
Radius :.
r2 = 9
r = 3 units
Example 3 :
(x + 3)2 + (y - 5)2 = 15
Solution :
The given equation of the circle is in the form of
(x - h)2 + (y - k)2 = r2
Center :
(h, k) = (-3, 5)
Radius :.
r2 = 15
r = √15 units
Example 4 :
x2 + y2 - 4x - 6y + 9 = 0
Solution :
The given equation of circle is in general form.
Comparing
x2 + y2 - 4x - 6y + 9 = 0
and
x2 + y2 + 2gx + 2fy + c = 0
we get
2g = -4 ----> g = -2 ----> -g = 2
2f = -6 ----> f = -3 ----> -f = 3
Center :
(-g, -f) = (2, 3)
Radius :
r = √(g2 + f2 - c)
= √(4 + 9 - 9)
= √4
= 2 units
Example 5 :
Find the center of the circle described on the line joining the points (1, 2) and (2, 4) as its diameter.
Solution :
Center of the circle = Midpoint of the diameter
Substitute (x1, y1) = (1, 2) and (x2, y2) = (2, 4).
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