Solving Problem with Circles in Backyard: Center (1,-1.5), Radius sqrt(29.25)

[Imperil]

In a backyard, there are two trees located at grid points A(-2, 3) and B(4, -6). The family cat is walking in the backyard. The line segments between the cat and the two trees are always perpendicular. Find the equation of the locus of the cat.

My Answer:

slope PA = (y - 3) / (x + 2)
slope PB = (y + 6) / (x - 4)
slope PA = -1 / slope PB (since the lines between PA and PB are perpendicular)

(y - 3) / (x + 2) = -1 / [(y + 6) / (x - 4)]
(y - 3) / (x + 2) = -1 * (x - 4) / (y + 6)
(y - 3) / (x + 2) = (-x+4) / (y + 6)

(y - 3) (y + 6) = (x + 2)(-x + 4)
y^2 + 3y - 18 = -x^2 + 2x + 8
x^2 + y^2 - 2x + 3y - 26 = 0

x^2 - 2x + 1 + y^2 + 3y + 2.25 = 26 + 1 + 2.25
(x - 1)^2 + (y + 1.5)^2 = 29.25

Therefore the cat is walking in a circle with center (1, -1.5) and radius sqrt(29.25).

I believe that my answer is incorrect but is there something I am missing? I have tried doing this question multiple times and I still can't find the correct answer.

EDIT: corrected a typo

 

[dx]

(y - 3) (y + 6) = (x + 2)(-x + 4)
y^2 + 3y - 18 = -x^2 + 2x + 8
x^2 + y^2 - 2x + 3y - 10 = 0

-18 - 8 = -26, not -10.

 

[Imperil]

Sorry yes that was a typo from a mistake I made on an old page.

I just checked by using x = 1 for a point on the circle and I believe that my answer is correct the way I have it (I edited the mistake)? I just tested with x=1 point on the circle and the lines now seem to be perpendicular

 

[dx]

Yes, your answer is correct.

 

[Imperil]

Thanks so much dx I really appreciate your time :)