Asked by Molly
Complete the square for this conic section in standard form:
x^2+y^2-6x+4y-12=0
x^2+y^2-6x+4y-12=0
Answers
Answered by
Reiny
x^2+y^2-6x+4y-12=0
x^2 - 6x + <b>9</b> + y^2 + 4y <b>4</b> = 12 + <b>9</b> + <b>4</b>
(x-3)^2 + (y+2)^2 = 25
x^2 - 6x + <b>9</b> + y^2 + 4y <b>4</b> = 12 + <b>9</b> + <b>4</b>
(x-3)^2 + (y+2)^2 = 25
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