(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!

Guest Nov 12, 2018

#1**+2 **

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

ElectricPavlov Nov 12, 2018

#1**+2 **

Best Answer

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

ElectricPavlov Nov 12, 2018

#2**0 **

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!

Guest Nov 12, 2018

edited by
Guest
Nov 12, 2018

edited by Guest Nov 12, 2018

edited by Guest Nov 12, 2018

#3**+2 **

I added 4 instead of r because that is what your posted question asked.

ElectricPavlov Nov 12, 2018