# How do you write the standard form of the following equation then graph x² + y² - 4x + 14y - 47 = 0 ?

We do a double completion of square--one with the

#1(x^2 - 4x + 4- 4) + 1(y^2 + 14y + 49 - 49) - 47 = 0#

#1(x^2 - 4x + 4) - 4 + 1(y^2 + 14y + 49) - 49 - 47 = 0#

#(x - 2)^2 - 4 + (y + 7)^2 - 49 - 47 = 0#

#(x -2)^2 + (y + 7)^2 = 100#

Now recall that when a circle is in the form

Therefore this is a circle with centre

Hopefully this helps!

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To write the standard form of the equation, complete the square for both (x) and (y) terms, then rearrange the terms. The standard form of the equation of a circle is ( (x - h)^2 + (y - k)^2 = r^2 ), where ( (h, k) ) is the center of the circle and ( r ) is the radius.

For the given equation (x^2 + y^2 - 4x + 14y - 47 = 0), complete the square for (x) and (y):

(x^2 - 4x + y^2 + 14y = 47)

(x^2 - 4x + 4 + y^2 + 14y + 49 = 47 + 4 + 49)

((x - 2)^2 + (y + 7)^2 = 100)

The equation is now in standard form. The center of the circle is at ((2, -7)) and the radius is (r = 10).

To graph the equation, plot the center at ((2, -7)) and draw a circle with radius (r = 10).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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