Maths (locus)

Circle : x² + y² - 10x + 6y + 21 = 0
Centre of the circle = (10/2, -6/2) = (5, -3)
Radius of the circle = √[(-10/2)² + (6/2)² - 21] = √13

Let (a, b) be the coordinates of the point of contact.

Slope of the tangent at (a, b) = 2/3
Slope of the normal at (a, b) :
(b + 3)/(a - 5) = -3/2 ...... [1]

Radius of the circle :
√[(a - 5)² + (b + 3)²] = √13 ...... [2]

From [1] :
b + 3 = -3(a - 5)/2 ...... [3]

Put [3] into [2] :
√{(a - 5)² + [-3(a - 5)/2)]²} = √13
(a - 5)² + [-3(a - 5)/2)]² = 13
(a - 5)² + (9/4)(a - 5)² = 13
(13/4)(a - 5)² = 13
(a - 5)² = 4
a - 5 = 2 or a - 5 = -2
a = 7 or a = 3

When a = 7 :
[3] : b + 3 = -3(7 - 5)/2
b + 3 = -3
b = -6

When a = 3 :
[3] : b + 3 = -3(3 - 5)/2
b + 3 = 3
b = 0

Hence, the coordinates of A and B are (7, -6) and (3, 0) respectively.