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X² + y² + 8x = 2x-20y-105; center​

User Ppeterka
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Answer:

To find the center of the given equation of a circle, we first need to rewrite it in the standard form:

x² + y² + 8x = 2x - 20y - 105

x² + 8x + y² = 2x - 20y - 105

Completing the square for the x terms:

(x² + 8x + 16) + y² = 2x - 20y - 105 + 16

(x + 4)² + y² = 2x - 20y - 89

Now we can see that the equation is in the standard form:

(x - (-4))² + (y - 0)² = r²

where the center of the circle is (-4, 0) and the radius squared is:

r² = 2x - 20y - 89

Note that the center of the circle can be found by taking the negative of the coefficients of x and y in the completed square terms and reversing their signs. In this case, the completed square term for x is (x + 4)², so the x-coordinate of the center is -4. The completed square term for y is y², so the y-coordinate of the center is 0.