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:
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Step-by-step explanation:
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Answer:
this is my attempt, it might not be right, so please dont be mad if wrong
Step-by-step explanation:
27 * x^3 = 6561x-5 , 49x + 1 = 117649x-2
Let
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