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Precalculus
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write the equation x^2-18y+y^2+14x+10=0. what is the center and radius of this circle
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\(x^2 - 18y + y^2 + 14x + 10 = 0\) Place all non-\(x\) terms on the right side \(x^2 + 14x = 18y - y^2 - 10\) Complete the square on \(x^2 + 14x\) by adding \(7^2\) to both sides: \(x^2 + 14x + 49 = 18y - y^2 - 10 + 49\) \((x+7)^2 = 18y - y^2 - 39\)
Correction: \((x + 7)^2 = 18y - y^2 + 39\)
Now Isolate \(y^2 - 18y\) \(y^2 - 18y = -(x + 7)^2 + 39\) Now complete the square on \(y^2 - 18y\) by adding 81 to both sides: \(y^2 - 18y + 81 = -(x + 7)^2 + 39 + 81\) \((x + 7)^2 + (y - 9)^2 = 120\)
The center of the circle would be (-7, 9) The radius of the circle would be \(\sqrt{120}\)
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