What is the center of the circle given by the equation (x + 5)2 + (y - 8)2 = 1?
Question
What is the center of the circle given by the equation
(x + 5)² + (y - 8)² = 1
?
Solution
The center of the circle given by the equation (x + 5)² + (y - 8)² = 1 is at the point (-5, 8).
Here's the step-by-step explanation:
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The general form of the equation of a circle is (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is the radius.
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In the given equation, (x + 5)² + (y - 8)² = 1, you can see that it's in the same form, but with a '+' before the '5' and a '-' before the '8'.
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This means that the 'h' value is -5 (because it's 'x - h', so 'x - (-5)' becomes 'x + 5') and the 'k' value is 8 (because it's 'y - k', so 'y - 8' gives 'k = 8').
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Therefore, the center of the circle is at the point (-5, 8).
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