# Is x^2+y^2=2y a circle?

• Dec 16th 2010, 07:21 PM
frankinaround
Is x^2+y^2=2y a circle?
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
x=0 y=0
x=1 y=1
x=0 y=2
x=-1 y=1
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=1 y=0
x=0 y=0.
x cant go anymore positive because y cant be negative in this equation.
• Dec 16th 2010, 07:37 PM
snowtea
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.
• Dec 16th 2010, 08:54 PM
Prove It
Yes $\displaystyle x^2 + y^2 = 2y$ is a circle...

$\displaystyle x^2 + y^2 - 2y = 0$

$\displaystyle x^2 + y^2 - 2y + (-1)^2 = (-1)^2$

$\displaystyle x^2 + (y-1)^2 = 1$.

So it's a circle of radius $\displaystyle 1$ unit, centred at $\displaystyle (0, 1)$.
• Dec 18th 2010, 01:28 PM
masoug
We can also say, $x^2+y^2=2y$ is an ellipse, but I guess that is superfluous...
• Dec 18th 2010, 01:30 PM
mr fantastic
Quote:

Originally Posted by masoug
We can also say, $x^2+y^2=2y$ is an ellipse, but I guess that is superfluous...

An ellipse with zero eccentricity. But then that would be like saying that a square is a rectangle, with all sides the same length. More confusing than helpful most of the time.
• Dec 18th 2010, 11:58 PM
frankinaround
Quote:

Originally Posted by snowtea
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.

Maybe its too late to be doing math... but I dont get it. your putting the line equation into the circle equation that way you know the point in the line equation has to be in the circle equation. which makes sense to me. Then your telling me to find the m's that give only 1 solution to x because if 1 m has 2 solutions to x then it must pass threw the circle twice which means its not tangent.

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.
• Dec 19th 2010, 12:17 AM
mr fantastic
Quote:

Originally Posted by frankinaround
Maybe its too late to be doing math... but I dont get it. your putting the line equation into the circle equation that way you know the point in the line equation has to be in the circle equation. which makes sense to me. Then your telling me to find the m's that give only 1 solution to x because if 1 m has 2 solutions to x then it must pass threw the circle twice which means its not tangent.

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.

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.
• Dec 19th 2010, 09:10 AM
frankinaround
Quote:

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.

I get as far as x^2 + (mx+4)(mx+2)=0 ...(1)
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.
• Dec 19th 2010, 09:42 AM
snowtea
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).
• Dec 19th 2010, 07:32 PM
frankinaround
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" ?
• Dec 19th 2010, 08:15 PM
mr fantastic
Quote:

Originally Posted by frankinaround
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" ?

I don't know why you are attempting this question if you have not learnt about solving quadratic equations (because if you have learnt this then you would surely know about the discriminant). I suggest you thoroughly review your class notes or textbook on this material. Or use Google.
• Dec 20th 2010, 11:49 AM
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.

$x^2+y^2=2y\Rightarrow\ x^2+y^2-2y=0\Rightarrow\ x^2+y^2-2y+1=1$

$(x-0)^2+(y-1)^2=1$ is a circle, centre $(0,1),$ radius $1.$

Tangents from $(0,4)$ to this circle create a pair of identical right-angled triangles as shown.

The hypotenuse is $4-1=3$ and another side is a radius of the circle
Pythagoras' theorem gives the 2nd of the perpendicular sides..

$3^2=1^2+8\Rightarrow$ the distance from $(0,4)$ to the tangent's point of contact

with the circle is $\sqrt{8}$

The slope of the positive-going tangent is $tan\alpha=\sqrt{8}$

so it's equation can be written using $y-4=\sqrt{8}(x-0)$

The 2nd tangent has slope $-tan\alpha=-\sqrt{8}$

For the algebraic solution, a line from $(0,4)$ 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 $(x,y)$ that is on both the circle and line if the line is a tangent.

$\displaystyle\ ax^2+bx+c=0\Rightarrow\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

This has no solution (line does not touch the circle) if $b^2-4ac<0$

It has one solution (line is tangent to the circle) if $b^2-4ac=0$

It has two solutions (line cuts through the circle) if $b^2-4ac>0$

You can see this just by examining the quadratic equation.

Where the $ax^2+bx+c=0$ quadratic equation comes from is through

substituting $y=mx+c$ into $x^2+y^2=2y$

since that will give you a quadratic in $x.$

We substitute to find the $x$ co-ordinate at which the tangent touches the circle.