can anyone help me please with this?

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

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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.....

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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

( 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.......

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

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