+0  
 
+1
121
1
avatar+281 

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
Sort: 

1+0 Answers

 #1
avatar+17601 
+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

geno3141  Jul 21, 2017

25 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details