center of circle with equation x^2+14x+y^2-6y+53=0
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Originally Posted by vegasgunner center of circle with equation x^2+14x+y^2-6y+53=0 $\displaystyle x^2+14x+49+y^2-6y+9=-53+49+9$
would it be (-7,3)??
Originally Posted by vegasgunner would it be (-7,3)?? You should know at least that much. Look up definitions,
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