+0  
 
+1
537
1
avatar+598 

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
avatar+17746 
+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( (x2 - x1)2 + (y2 - y1)2‚Äč )  =  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

40 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar
avatar
avatar