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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\$.

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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

Jul 21, 2017

### 1+0 Answers

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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

Jul 21, 2017