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sabi92  Jun 4, 2015

Best Answer 

 #3
avatar+78577 
+13

Here's the algebraic solution without using polar coordinates.....

( x + 6)^2 + ( y - 9)^2 = 52

x^2 + y^2  = 13   →  y^2  = 13 - x^2  →  y = ± √(13 - x^2)

 

Let us first guess that y = the positive root of √(13 - x^2)

 

Substitute this into the first equation.........

 

(x + 6)^2 + ( √(13 - x^2) - 9)^2 = 52       expand

 

x^2 + 12x  + 36 + 13 - x^2 - 18 √(13 - x^2) + 81 = 52     simplify

 

12x - 18 √(13 - x^2) = -78   divide through by 6

 

2x - 3 √(13 - x^2) = -13   rearrange

 

2x + 13  = 3√(13 - x^2)    square both sides

 

4x^2 + 52x + 169  = 9(13 - x^2)    simplify

 

4x^2 + 52x +169 = 117 - 9x^2     rearrange

 

13x^2 + 52x + 52  = 0       divide through by 13

 

x^2 + 4x + 4 = 0   factor

 

(x + 2)^2 = 0   take the square root of both sides

 

x + 2 = 0     so  x =  -2   and  y = √(13 - x^2)  = √(13 - (-2)^2) = √(13 - 4) = √9 = 3

 

And we have seen above that (-2, 3) is a solution

 

----------------------------------------------------------------------------------------------------

Now...let us assume that the negative root of  √(13 - x^2)   also might work for y

 

The algebra looks eerily similar to what we did before...........

 

(x + 6)^2 + ( -√(13 - x^2) - 9)^2 = 52

 

x ^2 + 12x + 36  + 13 - x^2 + 18√(13 - x^2) + 81  = 52

 

12x + 18√(13 - x^2) = -78     

 

2x + 3√(13 - x^2) = -13

 

2x + 13  = -3√(13 - x^2)   

 

4x^2 + 52x + 169  = 9(13 - x^2)

 

4x^2 + 52x + 169 = 117 - 9x^2

 

13x^2 + 52x + 52 = 0

 

x^2 + 4x + 4 = 0

 

(x + 2)^2  = 0  

 

x + 2  = 0   ...  so....    x = -2   and y = -√(13 - x^2)  =- √(13 - (-2)^2) =- √(13 - 4) = -√9 = -3

 

However...notice the problem in the first equation if y = -3

 

(-2 + 6)^2 + (-3 - 9)^2  =  4^2 + (-12)^2  =  16 + 144 =  160  and this does not equal 52  !!!

 

So....we only have one solution.....

 

--------------------------------------------------------------------------------------------------

 

See???......I told you it was messy  !!!

 

Let us offer up our many sacrifices to the Altar of Desmos, The Spirit of the Graphing Calculator

 

 

CPhill  Jun 4, 2015
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3+0 Answers

 #1
avatar+78577 
+10

( x + 6)^2 + ( y - 9)^2 = 52

x^2 + y^2  = 13  

 

The Algebra for this one could get a little messy......I might use a graphical approach....

 

https://www.desmos.com/calculator/cnkpzxaqrm

 

These are two circles that are tangent to each other....

 

The only "solution" point occurs at (-2, 3)....... And this is the point of tangency.......

 

 

CPhill  Jun 4, 2015
 #2
avatar+26326 
+10

Here's an algebraic approach (Chris is right - it's a little messy!):

simul eqns 1

 simul eqns 2

.

Alan  Jun 4, 2015
 #3
avatar+78577 
+13
Best Answer

Here's the algebraic solution without using polar coordinates.....

( x + 6)^2 + ( y - 9)^2 = 52

x^2 + y^2  = 13   →  y^2  = 13 - x^2  →  y = ± √(13 - x^2)

 

Let us first guess that y = the positive root of √(13 - x^2)

 

Substitute this into the first equation.........

 

(x + 6)^2 + ( √(13 - x^2) - 9)^2 = 52       expand

 

x^2 + 12x  + 36 + 13 - x^2 - 18 √(13 - x^2) + 81 = 52     simplify

 

12x - 18 √(13 - x^2) = -78   divide through by 6

 

2x - 3 √(13 - x^2) = -13   rearrange

 

2x + 13  = 3√(13 - x^2)    square both sides

 

4x^2 + 52x + 169  = 9(13 - x^2)    simplify

 

4x^2 + 52x +169 = 117 - 9x^2     rearrange

 

13x^2 + 52x + 52  = 0       divide through by 13

 

x^2 + 4x + 4 = 0   factor

 

(x + 2)^2 = 0   take the square root of both sides

 

x + 2 = 0     so  x =  -2   and  y = √(13 - x^2)  = √(13 - (-2)^2) = √(13 - 4) = √9 = 3

 

And we have seen above that (-2, 3) is a solution

 

----------------------------------------------------------------------------------------------------

Now...let us assume that the negative root of  √(13 - x^2)   also might work for y

 

The algebra looks eerily similar to what we did before...........

 

(x + 6)^2 + ( -√(13 - x^2) - 9)^2 = 52

 

x ^2 + 12x + 36  + 13 - x^2 + 18√(13 - x^2) + 81  = 52

 

12x + 18√(13 - x^2) = -78     

 

2x + 3√(13 - x^2) = -13

 

2x + 13  = -3√(13 - x^2)   

 

4x^2 + 52x + 169  = 9(13 - x^2)

 

4x^2 + 52x + 169 = 117 - 9x^2

 

13x^2 + 52x + 52 = 0

 

x^2 + 4x + 4 = 0

 

(x + 2)^2  = 0  

 

x + 2  = 0   ...  so....    x = -2   and y = -√(13 - x^2)  =- √(13 - (-2)^2) =- √(13 - 4) = -√9 = -3

 

However...notice the problem in the first equation if y = -3

 

(-2 + 6)^2 + (-3 - 9)^2  =  4^2 + (-12)^2  =  16 + 144 =  160  and this does not equal 52  !!!

 

So....we only have one solution.....

 

--------------------------------------------------------------------------------------------------

 

See???......I told you it was messy  !!!

 

Let us offer up our many sacrifices to the Altar of Desmos, The Spirit of the Graphing Calculator

 

 

CPhill  Jun 4, 2015

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