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morpeh [17]
4 years ago
5

Find the equation of the circle that passes through the point (-3,1) and has its center at C(-4,6)

Mathematics
1 answer:
qaws [65]4 years ago
5 0

Answer:

(x+4)^2 + (y-6)^2 = 29

Step-by-step explanation:

The center-radius form of the circle equation is in the format (x – h)^2 + (y – k)^2 = r^2, with the center being at the point (h, k)

Replacing the center C(-4,6):

(x+4)^2 + (y-6)^2 = r^2

then replacing the point (-3,1):

(-3+4)^2 + (1-6)^2 = r^2

1 + 25 = r^2

then the equation of the circle is:

(x+4)^2 + (y-6)^2 = 29

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Inflection point is the point where the second derivative of a graph is zero.

y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2

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I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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