Equation of circle
x^{2} + y^{2} = r^{2}
Center (0, 0) and radius is r.
(x - h)^{2} + (y - k)^{2} = r^{2}
Center (h, k) and radius is r.
x^{2} + y^{2} + 2gx + 2fy + c = 0
Center (-g, -f) and radius is r = √(g^{2} + f^{2} – c)
Write down the center and radius of each
circle below
Problem 1 :
x^{2} + y^{2} = 25
Solution :
The given equation of the circle is in the form of
x^{2} + y^{2} = r^{2}
So, the center of the given circle is (0, 0).
Radius :
r^{2} = 25
r = 5 units
Problem 2 :
x^{2} + y^{2} = 12
Solution :
The given equation of the circle is in the form of
x^{2} + y^{2} = r^{2}
So, the center of the given circle is (0, 0).
Radius :
r^{2} = 12
r^{ }= √12 units
Problem 3 :
(x – 3)^{2} + (y – 2)^{2} = 36
Solution :
The given equation of the circle is in the form of
(x – h)^{2} + (y – k)^{2} = r^{2}
Center :
(h, k) = (3, 2)
Radius :
r^{2} = 36
r = 6 units
Problem 4 :
(x + 1)^{2} + (y – 4)^{2} = 10
Solution :
The given equation of the circle is in the form of
(x – h)^{2} + (y – k)^{2} = r^{2}
Center :
(h, k) = (-1, 4)
Radius :
r^{2} = 10
r = √10 units
Problem 5 :
x^{2} + y^{2} – 10x – 6y - 2 = 0
Solution :
The given equation of circle is in general form.
Comparing
x^{2} + y^{2} – 10x – 6y - 2 = 0
Using the completing the square method,
x^{2} – 2 · x · 5 + 5^{2} - 5^{2} + y^{2} - 2 ·y·3 + 3^{2} - 3^{2} - 2 = 0
(x - 5)^{2} + (y - 3)^{2} - 25 - 9 - 2 = 0
(x - 5)^{2} + (y - 3)^{2} - 36 = 0
(x - 5)^{2} + (y - 3)^{2} = 36
Comparing with (x - h)^{2} + (y - k)^{2} = r^{2}
Center :
(-g, -f) = (5, 3)
Radius :
Radius = 6 units
Problem 6 :
x^{2} + y^{2} + 6x + 4y + 4 = 0
Solution :
The given equation of circle is in general form.
Comparing
x^{2} + y^{2} + 6x + 4y + 4 = 0
Using the completing the square method,
x^{2} + 2 · x · 3 + 3^{2} - 3^{2} + y^{2} + 2 ·y·2 + 2^{2} - 2^{2} + 4 = 0
(x + 3)^{2} + (y + 2)^{2} - 9 - 4 + 4 = 0
(x + 3)^{2} + (y + 2)^{2} - 9 = 0
(x + 3)^{2} + (y + 2)^{2} = 9
Center :
(-g, -f) = (-3, -2)
Radius :
Radius = 3 units
Problem 7 :
The point (a, 5) lies on the circle with equation x^{2} + y^{2} = 74. Find two values for a.
Solution :
Given, x^{2} + y^{2} = 74
The point (a, 5) = (x, y)
(a)^{2} + (5)^{2} = 74
(a)^{2} + 25 = 74
a^{2} = 74 – 25
a^{2} = 49
a = ±7
So, two values for a is 7, -7.
Problem 8 :
The point (3, c) lies on the circle x^{2} + y^{2} – 4x + 6y + 12 = 0. Find c.
Solution :
Given, x^{2} + y^{2} – 4x + 6y + 12 = 0
The point (3, c) = (x, y)
(3)^{2} + (c)^{2} – 4(3) + 6(c) + 12 = 0
9 + c^{2} – 12 + 6c + 12 = 0
9 + c^{2} + 6c = 0
c^{2} + 6c + 9 = 0
(c + 3) (c + 3) = 0
(c + 3)^{2} = 0
c + 3 = 0
c = -3
So, c is -3.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM