The center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)
<h3>Equation of a circle</h3>
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:
2
Step-by-step explanation:
Box in your 3m. Subtract 5 from both sides. You get 6. Now, divide 6 by 3 which is boxxed and you get 2.
Because otherwise the two circles you build wouldn't have any intersection!
You have to draw two circles, centered at the endpoints of the segment, and connect their intersection.
If the radii are less than half the length of the circle, the circles will have no interserction.
Least to greatest 52, 63, 55, 56, 62, 54, 59, 60, 53, 61, 88
BartSMP [9]
Answer:
52, 53, 54, 55, 56, 59, 60, 61, 62, 88
Step-by-step explanation:
You need to start from 52 to 88 (least to greatest) and then you will get answer.
Answer:
A number and its additive inverse are always the same distance from zero, and so they have the same absolute value.
Step-by-step explanation:
Hope this helps!!