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3 years ago

What is the equation of a circle with center (−8, 3) and radius 8?

Mathematics

1 answer:

Genrish500 [490]3 years ago

7 0

Answer:

x^2 + y^2 + 16x + 6y + 9 = 0

Step-by-step explanation:

Using the formula for equation of a circle

(x - a)^2 + (y + b)^2 = r^2

(a, b) - the center

r - radius of the circle

Inserting the values given in the question

(-8,3) and r = 8

a - -8

b - 3

r - 8

[ x -(-8)]^2 + (y+3)^2 = 8^2

(x + 8)^2 + (y + 3)^2 = 8^2

Solving the brackets

( x + 8)(x + 8) + (y +3)(y+3) = 64

x^2 + 16x + 64 + y^2 + 6y + 9 = 64

Rearranging algebrally,.

x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0

Bringing in 64, thereby changing the + sign to -

Therefore, the equation of the circle =

x^2 + y^2 + 16x + 6y + 9 = 0

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3 years ago

A. Find y in terms of x if ????y????x=x6y−7 and y(0)=4. y(x)= (8/7x^7+4^8)^(1/8) . B. For what x-interval is the solution define

topjm [15]

Answer:

Step-by-step explanation:

part A)

given

$x=x6y-7$

so y in terms of x would be

$y=(x+7)/x6$

part b)

in order to find intervals of function, we determine the intervals where the function has a positive first derivative . To find intervals, first we find the critical values, or points at which the first derivative of the function is equal to zero. For the given function the intervals would be

put zero in equation

$y=(x+7)/6x$

$y= infinity$

put 1

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put 2

$y= 0.75$

put 3

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hence we can see that the interval is from zero to positive infinity

(PINF)

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3 years ago