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True [87]
3 years ago
5

Pls help, easy geometry question

Mathematics
2 answers:
Lemur [1.5K]3 years ago
3 0

x^2 + y^2 + 2x - 6y + 1 = 0

x^2 +2x + 1 + y^2 - 6y = 0

x^2 + 2( x × 1) + 1^2 + y^2 - 2(y×3) + 3^2 -3^2 =0

(x+1)^2 + (y-3)^2 = 3^2

comparing with circle equation

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

where (a , b) is centre of circle.

so we get center of circle is

(-1, 3)

malfutka [58]3 years ago
3 0

Answer:

x²+y²+2x -6y + 1 = 0

x² +2x + 1 + y² - 6y = 0

x² + 2( x × 1) + 1² + y² - 2(y×3) + 3² -3² =0

(x+1)² + (y-3)² = 3²

The circle equation is:

(x-a)² + (y - b)² = r²

(a , b) = centre of circle.

So the centre of the circle is

<u>(-1, 3)</u>

Step-by-step explanation:

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Read 2 more answers
Hi can someone please help me?
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Both can be right because if the function goes up, turns and goes down, between x = -2 and x = 2, it can happen that  f(-2) = f(2) and then the average rate of change is [f(2) - f(-2)] / [2-(-2)] which is 0/4 = 0.
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