Distance from (-2,3) to (-1,1)

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The center of the circle defined by the equation x^2 + 2x + y^2 - 2y - 3 = 0 is correctly identified as (-1,1). The distance from the point (-2,3) to this center is calculated using the distance formula: √((-1+2)^2 + (1-3)^2). The calculation confirms the distance as √5. Additionally, the point (-2,3) is confirmed not to be on the circle.
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Homework Statement



Find the distance from the point (-2,3) to the centre of the circle x^2 + 2x + y^2 - 2y - 3 = 0


Homework Equations





The Attempt at a Solution



The circle, I calculated that the center point is (-1,1) is that correct?

Then, I just input it into the distance formula: √(-1+2)^2 + (1 - 3)^2

?
 
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nukeman said:

Homework Statement



Find the distance from the point (-2,3) to the centre of the circle x^2 + 2x + y^2 - 2y - 3 = 0

Homework Equations



The Attempt at a Solution



The circle, I calculated that the center point is (-1,1) is that correct?

Then, I just input it into the distance formula: √((-1+2)^2 + (1 - 3)^2 )

?
What's you question?

Is the point, (-2,3), on the circle? ... just asking?
 


nukeman said:

Homework Statement



Find the distance from the point (-2,3) to the centre of the circle x^2 + 2x + y^2 - 2y - 3 = 0


Homework Equations





The Attempt at a Solution



The circle, I calculated that the center point is (-1,1) is that correct?

Then, I just input it into the distance formula: √(-1+2)^2 + (1 - 3)^2

?
Yes, to both questions.
 

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