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
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( (x2 - x1)2 + (y2 - y1)2 ) = sqrt( (1 - -1)2 + (2 - 1)2 ) = sqrt( 22 + 12 ) = 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 - k2) = r2 ---> (x - -1)2 + (y - 1)2 = 5
---> (x + 1)2 + (y - 1)2 = 5
---> [ x2 + 2x + 1 ] + [ y2 - 2y + 1 ] = 5
---> x2 + y2 + 2x - 2y - 3 = 0
A = 1, B = 2, C = -2, D = -3
A + B + C + D = -2