The center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)
The standard equation of a circle is expressed as:
x^2 + y^2 + 2gx + 2fy + c = 0
where:
(-g, -f) is the centre of the circle
Given the equations
x^2 +y^2 – 12x – 2y +12 = 0
Compare
2gx = -12x
g = -6
Simiarly
-2y = 2fy
f = -1
Centre = (6, 1)
Hence the center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)
Learn more on equation of a circle here: brainly.com/question/1506955
Answer: 21599373.37618
Step-by-step explanation:
length: 10
width: 3
height: 1
10 x 3 x 1=30
Answer:
21,600
Detailed solution is given :