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X2 + y2 - 6x + 4y - 12 =0

1 Answer

2 votes

Answer:

y = sqrt(-x^2 + 6 x + 16) - 2 or y = -sqrt(-x^2 + 6 x + 16) - 2

Explanation:

Solve for y:

-12 - 6 x + x^2 + 4 y + y^2 = 0

Subtract x^2 - 6 x - 12 from both sides:

y^2 + 4 y = -x^2 + 6 x + 12

Add 4 to both sides:

y^2 + 4 y + 4 = -x^2 + 6 x + 16

Write the left hand side as a square:

(y + 2)^2 = -x^2 + 6 x + 16

Take the square root of both sides:

y + 2 = sqrt(-x^2 + 6 x + 16) or y + 2 = -sqrt(-x^2 + 6 x + 16)

Subtract 2 from both sides:

y = sqrt(-x^2 + 6 x + 16) - 2 or y + 2 = -sqrt(-x^2 + 6 x + 16)

Subtract 2 from both sides:

Answer: y = sqrt(-x^2 + 6 x + 16) - 2 or y = -sqrt(-x^2 + 6 x + 16) - 2

User Behnam Eskandari
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