To find equation of circle, we need two quantities.
i) Center of the circle
ii) Radius of the circle.
When we have these two information, using the formula given below, we can find equation of the circle.
x^{2} + y^{2} = r^{2}
Center is (0, 0) and radius is r.
(x - h)^{2} + (y - k)^{2} = r^{2}
Center is (h, k) and radius is r.
Write down
the equations of each of the circles shown below :
Problem 1 :
Solution :
Centre = (0, 0)
Radius r = 2
Equation of a circle is x^{2} + y^{2} = r^{2}
x^{2} + y^{2} = 2^{2}
x^{2} + y^{2} = 4
So, equation of a circle is x^{2} + y^{2} = r^{2}.
Problem 2 :
Solution :
Centre = (1, 0) and Radius r = 1
(h, k) = (1, 0)
Equation of a circle is (x – h)^{2} + (y – k)^{2} = r^{2}
(x – 1)^{2} + (y – 0)^{2} = 1
x^{2} + 1^{2} - 2(x)(1) + y^{2} = 1
x^{2} + 1 - 2x + y^{2} = 1
x^{2} + y^{2} - 2x + 1 = 1
x^{2} + y^{2} - 2x = 0
So, equation of a circle is x^{2} + y^{2} – 2x = 0.
Problem 3 :
Solution :
Centre = (3, 3) and Radius r = 3
(h, k) = (3, 3)
Equation of a circle is (x – h)^{2} + (y – k)^{2} = r^{2}
(x – 3)^{2} + (y – 3)^{2} = 3^{2}
x^{2} + 3^{2} - 2(x)(3) + y^{2} + 3^{2} - 2(y)(3) = 9
x^{2} + 9 – 6x + y^{2} + 9 - 6y = 9
x^{2} + y^{2} – 6x – 6y + 18 = 9
Subtract 9 from both sides.
x^{2} + y^{2} – 6x – 6y + 18 - 9 = 9 – 9
x^{2} + y^{2} – 6x – 6y + 9 = 0
So, equation of a circle is x^{2} + y^{2} – 6x – 6y + 9 = 0.
Problem 4 :
Solution :
Centre = (-1, -1) and Radius r = 1
(h, k) = (-1, -1)
Equation of a circle is (x – h)^{2} + (y – k)^{2} = r^{2}
(x + 1)^{2} + (y + 1)^{2} = 1^{2}
x^{2} + 1^{2} + 2(x)(1) + y^{2} + 1^{2} + 2(y)(1) = 1
x^{2} + 1 + 2x + y^{2} + 1 + 2y = 1
x^{2} + y^{2} + 2x + 2y + 2 = 1
x^{2} + y^{2} + 2x + 2y + 1 = 0
So, equation of a circle is x^{2} + y^{2} + 2x + 2y + 1 = 0.
Write down the equations of the following circles :
Problem 5 :
Solution :
Centre = (3, -1) and Radius r = 5
(h, k) = (3, -1)
Equation of a circle is (x – h)^{2} + (y – k)^{2} = r^{2}
(x – 3)^{2} + (y + 1)^{2} = 5^{2}
x^{2 } + 3^{2} – 2(x)(3) + y^{2} + 1^{2} + 2(y)(1) = 25
x^{2} + 9 – 6x + y^{2} + 1 + 2y = 25
x^{2} + y^{2} - 6x + 2y + 10 = 25
x^{2} + y^{2} - 6x + 2y - 15 = 0
So, equation of a circle is x^{2} + y^{2} - 6x + 2y - 15 = 0.
Problem 6 :
Solution :
Centre = (0, 6) and Radius r = 6
(h, k) = (0, 6)
Equation of a circle is (x – h)^{2} + (y – k)^{2} = r^{2}
(x – 0)^{2} + (y - 6)^{2} = 6^{2}
x^{2 } + 0^{2} – 2(x)(0) + y^{2} + 6^{2} - 2(y)(6) = 36
x^{2} + 0 – 0x + y^{2} + 36 - 12y = 36
x^{2} + y^{2} - 12y + 36 = 36
x^{2} + y^{2} - 12y = 0
So, equation of a circle is x^{2} + y^{2} - 12y = 0.
Problem 7 :
Solution :
Centre = (4, 5) and Radius r = 4
(h, k) = (4, 5)
Equation of a circle is (x – h)^{2} + (y – k)^{2} = r^{2}
(x – 4)^{2} + (y - 5)^{2} = 4^{2}
x^{2 } + 4^{2} – 2(x)(4) + y^{2} + 5^{2} - 2(y)(5) = 16
x^{2 } + 16 – 8x + y^{2} + 25 - 10y = 16
x^{2 } – 8x + y^{2} - 10y + 16 - 16 + 25 = 0
x^{2 } – 8x + y^{2} - 10y + 25 = 0
So, equation of a circle is x^{2} + y^{2} – 8x - 10y + 25 = 0.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM