If someone could CORRECTLY help me on this please

We are given:

A  =  (4, 1)

B  =  (6, 2)

C  =  (-1, 2)

And we can let:

P  =  (x, y)

By the Pythagorean Theorem/distance formula we can say:

PA²  =   (x - 4)²  +  (y - 1)²

PB²  =   (x - 6)²  +  (y - 2)²

PC²  =   (x + 1)²  +  (y - 2)²

Then....

PA² + PB² + PC²   =   (x - 4)²  +  (y - 1)²   +   (x - 6)²  +  (y - 2)²   +   (x + 1)²  +  (y - 2)²

Expand each term, then combine like terms to get:

PA² + PB² + PC²   =   3x²  -  18x  +  3y²  -  10y  +  62

(I used this to do that because I am lazy.)

Next we want to get the expression on the right side of the equation into the form:

3PQ²  +  k

which is:

3[ (x - something)²  +  (y - something)² ] + k

To do that, we need to complete the squares of the x terms and the y terms.

PA² + PB² + PC²   =   3(x²  -  6x)  +  3(y²  -  \(\frac{10}{3}\)y)  +  62

PA² + PB² + PC²   =   3(x²  -  6x + 9 - 9)  +  3(y²  -  \(\frac{10}{3}\)y + \(\frac{25}{9}\)  -  \(\frac{25}{9}\))  +  62

PA² + PB² + PC²   =   3( (x - 3)² - 9)  +  3( (y - \(\frac53\))²  -  \(\frac{25}{9}\))  +  62

PA² + PB² + PC²   =   3(x - 3)² - 27  +  3(y - \(\frac53\))²  -  \(\frac{25}{3}\)  +  62

PA² + PB² + PC²   =   3(x - 3)²  +  3(y - \(\frac53\))²  -  \(\frac{25}{3}\)  +  62 - 27

PA² + PB² + PC²   =   3[ (x - 3)²  +  (y - \(\frac53\))² ]   -  \(\frac{25}{3}\)  +  62 - 27

PA² + PB² + PC²   =   3[ (x - 3)²  +  (y - \(\frac53\))² ]   +   \(\frac{80}{3}\)

(Check)

Now it is in the desired form, and so we can pick out that    k   =   \(\frac{80}{3}\)

BTW, I came across this answer. It works out a bit more cleanly if  A  =  (4, -1).  Just wanted to mention this in case that is what you meant.