- #1
xxmegxx
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What is the equation of the circle with a center point of (10, -14) when the circle is tangent to x=13?
D = √(13-10)^2 + (0-(14))^2
D = √(3)^2 + (14))^2
D = √9+196
D = √205
Radius = √205
(x-10)^2 + (y-(-14))^2 = √205^2
(x-10)^2 + (y+14)^2 = 205
But how am I suppose to graph this?
D = √(13-10)^2 + (0-(14))^2
D = √(3)^2 + (14))^2
D = √9+196
D = √205
Radius = √205
(x-10)^2 + (y-(-14))^2 = √205^2
(x-10)^2 + (y+14)^2 = 205
But how am I suppose to graph this?
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