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The equation of the circle shown in the following diagram can be written as $x^2 + Ay^2 + Bx + Cy + D = 0$. Find $A+B+C+D$.

Center of circle is (-1,1), point on circle is (1,2)

diagram dont know how to get onto web2.0calc.com

michaelcai Jul 21, 2017

#1**+2 **

Since the center of the circle is (-1,1) and a point on the circle is (1,2), the radius of the circle can be found by finding the distance from the center to the point on the circle.

distance = sqrt( (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} ) = sqrt( (1 - -1)^{2} + (2 - 1)^{2} ) = sqrt( 2^{2} + 1^{2} ) = sqrt( 4 + 1) = sqrt( 5 )

If the center of the circle is (h, k) and the radius is r, the tormula for the equation of the circle:

(x - h)^{2} + (y - k^{2}) = r^{2} ---> (x - -1)^{2} + (y - 1)^{2} = 5

---> (x + 1)^{2} + (y - 1)^{2} = 5

---> [ x^{2} + 2x + 1 ] + [ y^{2} - 2y + 1 ] = 5

---> x^{2} + y^{2} + 2x - 2y - 3 = 0

A = 1, B = 2, C = -2, D = -3

A + B + C + D = -2

geno3141 Jul 21, 2017