+0  
 
0
61
2
avatar

The circle centered at $(2,-1)$ and with radius $4$ intersects the circle centered at $(2,5)$ and with radius $\sqrt{10}$ at two points $A$ and $B$. Find $(AB)^2$.

 Feb 11, 2020
 #2
avatar+109740 
+1

Equation of first circle  :

(x - 2)^2   + ( y + 1)^2  = 16

Equation of second circle :

(x - 2)^2   + (y - 5)^2  = 10

 

Subtract  the second equation from the first and we have

 

(y + 1)^2  - ( y - 5)^2  =  6           simplify

 

y^2 + 2y + 1  -  y^2  + 10y - 25  =  6

 

12y - 24  =  6

 

12y =  30

 

y = 30/12 =  2.5 

 

Using the first  equation to find the x intersection points, we have  that

 

(x - 2)^2  + ( 2.5 + 1)^2 = 16

 

(x - 2)^2  + (3.5)^2   =16

 

(x - 2)^2   +  12.25  =16

 

(x -2)^2  =  3.75

 

(x - 2)^2  = 15/4       take both roots

 

 x - 2 =  ±√15/ 2

 

x =  ±√15/2  + 4/2

 

x  =  [ 4 +√15] / 2         and    x =  [ 4 - √15] / 2

 

The   distance between  these points  =   [ 4 + √15 ] / 2  - [ 4 - √15] / 2   =  2√15/ 2  =  √15

 

So.....the square of this distance =  15

 

 

 cool cool cool

 Feb 11, 2020

53 Online Users

avatar
avatar
avatar
avatar
avatar