A-maths!!

a
將A(6,0)代入x^2+y^2-4x+y-12=0
LHS
= 6^2 + 0^2 - 4(6) + 0 - 12
= 0
=RHS
∴A(6,0)位於圓x^2+y^2-4x+y-12=0上
將B(0,-4)代入x^2+y^2-4x+y-12=0
LHS
= 0^2 + 4^2 - 4(0) - 4 - 12
= 0
= RHS
∴B(0,-4)位於圓x^2+y^2-4x+y-12=0上

b
通過A點的圓切線:
6x + 0y - 4(6+x)÷2 + (0+y)÷2 - 12 = 0
12x - 24 - 4x + y -24 = 0
8x + y - 48 = 0---------------------------(1)

通過B點的圓切線:
0x + (-4)y - 4(0+x)÷2 + (-4+y)÷2 - 12 = 0
-8y - 4x -4 +y - 24 = 0
-12x + y - 28 =0-------------------------(2)

(1) - (2): 20x - 20 = 0
x = 1
將 x=1代入(2):
-12 + y - 28 =0
y = 40

∴交點是(1,40)

For part (a),
you can simply substitute the coordinates of A and B into the equation of the circle.
For point A,
(6)^2+(0)^2-4(6)+(0)-12=0
For point B,
(0)^2+(-4)^2-4(0)+(-4)-12=0
So point A and point B lie on the circle.

But for part B, I don't really understand the question,
so I cannot answer that...
=,=