(No explanation needed, but if you do, THANK YOU SO MUCH! :) )
The equation of a circle which has a center at (-5,2) can be written as $Ax^2 + 2y^2 + Bx + Cy = 40$. Let $r$ be the radius of the circle. Find $A + B + C + 4$.
(I tried to do it, and I got $14 + 7 sqrt 2$, but apparently, that is incorrect.)
Thank you so much!
+36916 Cirle formula
(x-h)^2 + (y-k)^2 = r^2 center = h,k
( x+5)^2 + (y-2)^2 =r^2
x^2 + 10x + 25 + y^2 -4y + 4 = r^2
x^2 + y^2 + 10x -4y = r^2 -29 Multiply both sides by 2 to get the 2y^2
2 x^2 + 2y^2 + 20x - 8y = 2r^2 - 58 and 2^r^2 - 58 = 40 2r^2 = 98 r^2 = 49 r= 7
A+B + C +4 = 2 + 20 + (-8) + 4 = 18
IF you meant A+B+C+r then = 2+20+(-8) + 7 = 21
+36916 Cirle formula
(x-h)^2 + (y-k)^2 = r^2 center = h,k
( x+5)^2 + (y-2)^2 =r^2
x^2 + 10x + 25 + y^2 -4y + 4 = r^2
x^2 + y^2 + 10x -4y = r^2 -29 Multiply both sides by 2 to get the 2y^2
2 x^2 + 2y^2 + 20x - 8y = 2r^2 - 58 and 2^r^2 - 58 = 40 2r^2 = 98 r^2 = 49 r= 7
A+B + C +4 = 2 + 20 + (-8) + 4 = 18
IF you meant A+B+C+r then = 2+20+(-8) + 7 = 21
Thank you so much, Electric Pavlov!
(I checked your work and saw r = 7, but when you added A, B, C, and r, you put r as 4. I think you meant to put r as 7, so the answer was supposed to be 21.) Thank you anyways, I really appreciate the response.
Sorry, I just realized that I typed my question wrong! :) But thank you anyways!
+36916 I added 4 instead of r because that is what your posted question asked. 
