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Find the equation in standard form for the circle whose diameter has an endpoint at (-3, -4) and the origin. 4 x 2 + 4 y 2 - 12 x + 16 y = 0 4 x 2 + 4 y 2 + 12 x - 16 y = 0 4 x …

Find the equation in standard form for the circle whose diameter has an endpoint at (-3, -4) and the origin. 4 x 2 + 4 y 2 - 12 x + 16 y = 0 4 x 2 + 4 y 2 + 12 x - 16 y = 0 4 x …
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Find the equation in standard form for the circle whose diameter has an endpoint at (-3, -4) and the origin.

4 x 2 + 4 y 2 - 12 x + 16 y = 0
4 x 2 + 4 y 2 + 12 x - 16 y = 0
4 x 2 + 4 y 2 - 12 x - 16 y = 0
4 x 2 + 4 y 2 + 12 x + 16 y = 0

User Jassi
by
8.4k points

1 Answer

4 votes

The answer is option A.


4 x 2 + 4 y 2 - 12 x + 16 y = 0

User TomRoche
by
7.9k points
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