coordinate geometry

Let's get an equation for the line....we have that

y  = -2(x - 3) + 5         simplify

y = -2x + 6 + 5

y  =  -2x + 11        (1)

We can let  (3, 5)  be the center of a circle   with a radius  of  AB

So  we have the circular  equation

(x  - 3)*2  + ( y - 5)^2  =  (6√5)^2      sub (1)  into (2)  for  y   and we have that

(x - 3)^2  + ( -2x + 11 -  5)^2  =   180      simplify

(x -3)^2  + ( 6 - 2x)^2  = 180

x^2 - 6x + 9  + 4x^2 - 24x + 36   = 180

5x^2 - 30x + 45  = 180

5x^2 -30x - 135  =  0         divide through by 5

x^2  - 6x  - 27   =   0      factor

( x - 9) ( x + 3)  = 0

Set  both factors to 0 and solve for x and we have that

x =  9        and   x   =   -3

Using the equation of the line

y  = -2(9 -3) + 5              y  = -2(-3 - 3) + 5

y  = -12 + 5                    y  =  12 + 5

y  =  - 7                           y  = 17

So....the two possible endpoints for  B  are   ( -3, 17)   and  ( 9, -7)

Here's a graph   :    https://www.desmos.com/calculator/viyvhipurq

A little different approch....

same method to find the line    y = -2x +11

Now use the distance formula    d²= (x1-x2)^2 +(y1-y2)^2           d^2 = (6 sqrt5)^2 = 180

Using the point given (3,5) as x2    and y₂ = -2x+11 as shown above

(x-3)^2 + (y- (-2x+11)^2 = 180   

x^2 + 6x + 9 +4x^2 -24x +36     -180  = 0

5x^2-30x -135 = 0          Use quadratic formula to find x = 9      or      -3

Sub these values into the red line equation to find y = -7    and 17                           (9,-7)    and (-3,17)   

thanks, I like both of these methods, I struggle with this topic so I will change the numbers and use both to see what I like better, I will post later asking if it is right or not.