Substitue y=mx+4 into the circle equation
x^2 + (mx+4)^2 = 2(mx+4)
Find all m that gives exactly 1 solution for x. You can use completing the square for x to get the formula in the form (x - A)^2 = B.
This has a unique solution for x iff B = 0.
Find the equations of all lines that are tangent to the circle x^2+y^2=2y and pass threw the point (0,4). Hint : the line y=mx+4 is tangent to the circle if it intersects the circle at 1 point.
I cant solve this problem. I put my work below but trying to understand my mistakes really might be a waist of time for whoevers reading this. A summery of all my work below is I cant figure out how to make x^2+y^2=2y into a circle. (also the tangent part of the equation looks hard to.)
I dont see how x^2 +y^2=2y is a circle. Is (0,0) the origin? I tried mapping it but if center is origin then y cant be negative. I got :
(x-0)^2 + (y-0)^2 = 2y
But y cant be negative because the left side of equation cant be negative.
Also I realize that square root of 2y cant be radius.
BUT x^2 + (y-1)^2=1 gives me
x cant go anymore positive because y cant be negative in this equation.
OK. So you mean complete the square of x from x^2 + (mx+4)^2=2(mx+4). Im not sure what you mean by this though.
I get x^2 + mx^2 + 8mx +16 = 2mx + 8
I can complete the square for mx^2 + 8mx but why would that help me? (x+1)(x-1)+(mx+3)^2=0. I dunno im confused as to what im completing the square of exactly and why (the mx or just the x^2) and as to how to find the m's to get 1 solution of x. However its 3am, i doubt id have much luck after 8 hours sleep either haha.
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.
where does b^2-4ac come from? I sort of get that you need to find all the m's (and maybe x's) that make equation (1) = 0
but I dont see how to turn (1) into an equation to do this. Im operating now on the level of trying really hard to understand, as opposed to understand. I guess.
Never mind about completing the square.
What Mr. Fantastic said is easier.
x^2 + (mx+4)^2 = 2(mx+4)
you simplified to:
x^2 + mx^2 + 8mx +16 = 2mx + 8
now simplify to:
(m+1)x^2 + 6mx + 8 = 0
a = m+1
b = 6m
c = 8
Set: b^2 - 4ac = 0
b^2 - 4ac = (6m)^2 - 4(m+1)(8) = 36m^2 - 32m - 32 = 0
9m^2 - 8m - 8 = 0
Now solve for m (assuming I did not make some silly mistake).
I get why you put the line equation into the circle equation. But why do you need to get it to ax + bx + c to get b^2-4ac = 0? why does b^2-4ac=0 solve the equation?
mr fantastic said : "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."
Ok, I actually dont know that... Any ideas on where I can go to learn about that? "discriminants" ? why does the answer to equation (3) (ax^2+bx+c) depend on the value of the "discriminant" ?
You could work out the geometry,
but the question is surely asking you to make the transition
to working out the solution without a compass and ruler.
is a circle, centre radius
Tangents from to this circle create a pair of identical right-angled triangles as shown.
The hypotenuse is and another side is a radius of the circle
Pythagoras' theorem gives the 2nd of the perpendicular sides..
the distance from to the tangent's point of contact
with the circle is
The slope of the positive-going tangent is
so it's equation can be written using
The 2nd tangent has slope
For the algebraic solution, a line from could miss the circle
or be a tangent to the circle,
or cross the circle twice.
If you are finding points of contact between the circle and line,
you are looking for a single that is on both the circle and line if the line is a tangent.
This has no solution (line does not touch the circle) if
It has one solution (line is tangent to the circle) if
It has two solutions (line cuts through the circle) if
You can see this just by examining the quadratic equation.
Where the quadratic equation comes from is through
since that will give you a quadratic in
We substitute to find the co-ordinate at which the tangent touches the circle.