Originally Posted by

**mr fantastic** As has been mentioned earlier, the solution to the simultaneous equations

x^2+y^2=2y ... (1)

y = mx + 4 .... (2) (which is the general equation of a line that passes through the point (0,4) and you should have been able to get this without the hint)

is found by substituting (2) into (1):

x^2 + (mx+4)^2 = 2(mx+4).

You must expand and simplify this equation so that it has the form

ax^2 + bx + c = 0 ... (3)

where a, b and c obviously depend on m.

You should know that the number of solutions to equation (3) depends on the value of the discriminant b^2 - 4ac. So my advice is to get an expression for the discriminant, set it equal to zero (why?) and solve for m.