1. The equation of a circle is given by x^2+y^2-10x-8y+16=0. Find the x-intercept(s) and the y-intercept(s) of the circle.

2. The equation of a circle is given by x^2+y^2-6x-4y-12=0. Find the equation of the straight line joining the orgin to the circle.

3. Find the equation of the diameter passing through the point (5,1) on the circle x^2+y^2+4x-2y-11=0.

4. One end of a diameter of the circle C: x^2+y^2-6x-4y-12=0 is(-1,-1). Find the coordinates of the other end.

1/ Sub in x=0 to find the y-intercepts and y=0 to find the x-intercepts.  Remember that you get two solutions when taking square roots.

2/ Um... there are many straight lines that join the origin to "the circle".  Assuming that you meant the centre of the circle, you can rewrite the circle equation like this:
x^2+y^2-6x-4y-12=0
(x^2-6x) + (y^2-4y) - 12 = 0.  Since (6/2)^2 = 9 and (4/2)^2 = 4, we can write:
(x^2-6x+9)-9 + (y^2-4y+4)-4 - 12 = 0
(x-3)^2 + (y-2)^2 = 9+4+12.  The brackets tell you that the centre is (3, 2).
Then, let (x1, y1) = (0, 0), and (x2, y2) = (3, 2), and sub into (y-y1)/(x-x1) = (y2-y1)/(x2-x1), the rearrange to get the equation.

3/ Use the same method as before find the centre of the circle.  Then, sub in this point and (5, 1) into the formula above.

4/ Use the same methods as before to find the centre of the circle.  Note the increase/decrease in x and y values required to move from (-1, -1) to the centre.  Repeat that move starting from the centre, and you will get the other point.

2.呢度唔係好明
   x^2+y^2-6x-4y-12=0
  (x^2-6x) + (y^2-4y) - 12 = 0.