SOLUTION: find the area enclosed by the curve x^2+y^2-3x-8y+18=0

Question 1143791: find the area enclosed by the curve x^2+y^2-3x-8y+18=0
Answer by ikleyn(52786) (Show Source): You can put this solution on YOUR website!
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The curve is a circle, and all you need is to find its radius.


For it, transform the given equation of a circle from its general form to its standard form.


For it, complete the squares for x- and y-terms separately in the given equation


    x^2 + y^2 - 3x - 8y + 18 = 0

    (x^2 - 3x) + (y^2 - 8y) = -18

    (x^2 - 3x + 2.25) + (y^2 - 8y + 16) = -18 + 2.25 + 16

     +  = 0.25.


Thus you have the circle centered at the point (1.5,4) and having the radius of  r =  = 0.5.


The area of this circle is   =  = .    ANSWER

Solved.

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