A circle has a radius of 6/7 units and is centered at (-2.3,0) What is the equation of the circle

Mathematics

1 answer:

N76 [4]3 years ago

7 0

Answer:

(x+2.3)^2 + (y) ^2 = (6/7)^2

Step-by-step explanation:

The equation of a circle can be written as

(x-h)^2 + (y-k) ^2 = r^2 where ( h,k) is the center and r is the radius

(x- -2.3)^2 + (y-0) ^2 = (6/7)^2

(x+2.3)^2 + (y) ^2 = (6/7)^2

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Answer:

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Step-by-step explanation:

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natta225 [31]

Answer:

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Step-by-step explanation:

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Given

Consider a= α and b= β

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Putting values of αβ and α+β

$(\frac {-4}{k})^2 -2( \frac {4}{k}) = 24\\\frac {16}{k^2} - \frac {8}{k} = 24\\Multiplying\,\, the \,\, equation\,\, with\,\, 8K^2\\ 2 - k= 3K^2\\3k^2-2+k=0\\or\\3k^2+k-2=0\\3k^2+3k-2k-2=0\\3k(k+1)-2(k+1)=0\\(3k-2)(k+1)=0\\3k-2=0 \,\,and\,\, k+1 =0\\k= 2/3 \,\,and\,\, k=-1$