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Q = (1,5) and k = 18.

Thanks

10x = 180

X^2 + y^2 - 8x + 2y = - 1    'complete the square for x and y

(x-4)^2   +  (y+1)^2  = -1 +16 + 1  = 16 

(x-4)^2   + ( y+1)^2 = 16  

    Do you recognize the standard form of a circle equation?    center h, k   and r^2 are looking right at ya !  

Thank you both!

There IS a formula for point to line distance...but I am not going to use it

slope of given line is -1    perpindicular is  -  1 / -1  = 1

point slope form going thru 7,-6   and perpindicular

 y+6 = 1 (x-7)     or    y = x-13

the intersection of these two lines is   found:     x - 13 = -x -3     so x = 5      then y = -8

Now distance between   7, -6    and    5, -8

        (7-5)^2  + (-6- -8)^2 = d^2

d = sqrt 8 = 2 sqrt2

Dragan's answer is this:   YZ = 9.428571429 = 66 / 7  

\(4!\cdot2!\cdot5!\cdot3!\implies24\cdot2\cdot120\cdot6\implies34560\)

I got three solutions:

1. \(6\)

2. \(\frac {3+i\sqrt3}2\)

3. \(\dfrac {3-1\sqrt3}2\)

You can also write it as 

\(6, \dfrac {3\pm i\sqrt3}2\)

Yes, but there are 2 more that you can't find by bashing it.

uhm no. your not even close lmao.

Much swaver approach than my bash method. 

Slope of given line = -1/4

slope of new line = -1/4

in point slope form   (to go thru 1, -10)    y+10 = -1/4 (x-1)

                                    re-arrange to slope intercept form         y = -1/4x - 9 3/4

Oh, sorry about being rash about it.

Slopes are equal:

\(y=-\frac 14 x + c\)

Success! \(y=-\frac14x-9.75\)

I am sorry, but that is incorrect, very close, however. (I don't know the actual ans., but based on Dragan's decimal, I can tell you are close.)

Q = (-2,7) and k = 44.

What it means is this:

Message me for clarification if needed.

sorry man i know  and i did so i was about to come back say nvm but yea thanks man...

Just plug in again using the strategy I JUST showed you:

\(g(4)=3(4)+1 = \mathbf{13}\\\\\implies f(13)=13^2+13\implies169+13\implies \color {blue}{\mathbf{\boxed {182}}}\)