A(3,0) and B(-3,0) are the two fixed points Find the locus of a point P at which AB subtend a right angle at P

Step-by-step explanation:

Let's take a general point P(x,y). If there is a right angle at P then the slopes of the lines passing through AP and BP must be perpendicular.

Two lines of slopes m1 and m2 are perpendicular if:

m1*m2=-1

The slope of the line passing through P(x,y) and A(3,0) is:

The slope of the line passing through P(x,y) and B(-3,0) is:

Substituting in the equation:

Operating:

The locus is a circle centered at the origin with radius 3

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm