(a) I assume you mean (x-8)²+(y-4)²=100.
If the distance between (-6,6) and the centre (8,4) is greater than the radius 10, the point is outside the circle.
This distance is √((8-(-6))²+(4-6)²)=√(14²+2²)=√200=10√2. This is greater than 10, so the point is outside the circle.
(b) I assume you mean the tangent from the given point (-6,6) to the circle.
Differentiating the equation of the circle :
2(x-8)+2(y-4)dy/dx=0, so dy/dx=(8-x)/(y-4) at the point (x,y).
The slope from (x,y) to the point (-6,6) has slope (y-6)/(x+6).
These slopes must be equal so:
(8-x)/(y-4)=(y-6)/(x+6),
(8-x)(x+6)=(y-6)(y-4),
2x+48-x²=y²-10y+24,
x²+y²-2x-10y-24=0,
(x-1)²+(y-5)²=50, which is a circle centre (1,5).
Subtract this from the equation of the circle and employ the difference of two squares:
(x-8-x+1)(x-8+x-1)+(y-4-y+5)(y-4+y-5)=50,
-7(2x-9)+2y-9=50,
-14x+63+2y-9=50,
-14x+2y=-4, 7x-y=2, y=7x-2.
Substitute for y:
(x-1)²+(7x-7)²=50,
x²-2x+1+49x²-98x+49=50,
50x²-100x=0=50x(x-2), so x=0 or 2.
y=7x-2=-2 or 12.
Therefore the circles intersect at (2,12) and (0,-2) (the tangent points).
The slopes of the tangent lines from the point (-6,6) are:
(12-6)/(2+6)=6/8=3/4; (6+2)/-6=-8/6=-4/3.
The length of the tangent line is √(6²+8²)=√100=10 which is the same as the radius of the given circle. Therefore the triangle formed by joining the point (-6,6) to the centre of the circle (8,4) is a right isosceles triangle, making the angles 45°.
(c) The equation of the line is y-6=3(x+6)/4=3x/4+9/2, y=3x/4+21/2.
(d) The question is not clear—coordinates of what? If you mean the intercepts, then the x-intercept is when y=0: 3x/4=-21/2, x=-(4/3)(21/2)=-14; and the y-intercept is 21/2. Intercepts are (-14,0) and (0,21/2).