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(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!

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

Nov 12, 2018
edited by ElectricPavlov  Nov 12, 2018
edited by ElectricPavlov  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
edited by ElectricPavlov  Nov 12, 2018
edited by 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
#3
+2

I added 4 instead of r   because that is what your posted question asked. Nov 12, 2018