The Locus of Q consider point A and B Consider two fixed points A(0,-2) and B(0,4) on a rectangular coordinate plane.
A moving point Q such that QA is always perpendicular to QB.
I want to know how would the graph be like and the equation of the Locus of Q?
Thank you!
 A: the locus of $Q$ is a circle whose diameter is $AB$you derive the equation if you take $Q=(x,y)$ 
(1) find the slopes of $QA$ and $QB$ by using the formula $rise/run.$ 
(2) you also need to use that the product of slopes of two perpendicular lines is negative one.
see if you can do these two steps. 
your first comment is right. in the second one you forgot to cross multiply the $x^2$  and $-1$
$$-1 = {(y-(-2) \over (x - 0)} {(y - 4) \over( x - 0)} = {y^2 - 2y-8 \over x^2}$$ which simplifies to $$x^2 + y^2 - 2y + 8 = 0 \text { or  } x^2 + (y-1)^2 = 9.$$ which is a circle centered at (0,1) and radius 3 units.
A: If u will plot the points A and B on the cartesian plane and try to locate Q such that QA is perpendicular to QB, u can see that it's a circle.
 Its obvious because angle in a semicircle is always a right angle and Q will lie on a circle with AB as its one of the diameters. The centre will be the midpoint of A and B that is
C(0, 1) and radius will be AC=BC=3 units.
So concluding
The locus is a circle centered at (0, 1) with radius 3.
