+0  
 
+1
357
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

michaelcai  Jul 21, 2017
 #1
avatar+17743 
+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

23 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.