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Need help on a level maths question I've been stumped on for a while

The question is find the values of m such that y = mx is a tangent to (x-1)^2 + y^2 = 0.8

I tried solving it simultaneously but I only had one answer which isn't right as there are mean to be 2 answers.
I know the centre of the circle is (1,0)
I know the radius is 0.64
But the m is really making life harder.
Can anyone help me?
Pls...
Original post by BlueAgent214
The question is find the values of m such that y = mx is a tangent to (x-1)^2 + y^2 = 0.8

I tried solving it simultaneously but I only had one answer which isn't right as there are mean to be 2 answers.
I know the centre of the circle is (1,0)
I know the radius is 0.64
But the m is really making life harder.
Can anyone help me?
Pls.

Hi, isn't the radius root 0.8 not 0.8^2?
What did you get as your x and y value for solving?
Reply 3
Avatar for Coder214
Coder214
OP
9
Original post by DennisChin
Hi, isn't the radius root 0.8 not 0.8^2?

Oh yh. Sorry for that mistake
Reply 4
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Coder214
OP
9
Original post by DennisChin
What did you get as your x and y value for solving?

That's the problem. I don't know how to with the m in the way
I'll start you off, draw a radius line (from centre to tangent point).
Reply 6
Avatar for simon0
simon0
13
You have two equations:
1) A curve (in this case a circle): (x−1)2+y2=0.8, (x-1)^{2} + y^{2} = 0.8,
2) A straight line: y=mx. y = mx.

At the point where the two curves meet, the x and y coordinates must be the same.
One method is to substitute one equation into another.

Spoiler


You should end up with a quadratic you can set to zero.
If you drew the diagram, how many times does the tangent touch the circle?
Then how does this relate to the number of solutions to the quadratic?

------------------------------------------------------------------------------------------------

As a separate point, for the radius of the circle, using the equation:

r2=0.8. r^{2} = 0.8.

Then what is the radius?

-----------------------------------------------------------------------------------------------------


Tell us if you need more help.
(edited 4 years ago)
Reply 7
Avatar for Coder214
Coder214
OP
9
Thanks everyone for your help. I got m = 2 and -2.
There are two answers cos there is a power of 2
Original post by BlueAgent214
The question is find the values of m such that y = mx is a tangent to (x-1)^2 + y^2 = 0.8

I tried solving it simultaneously but I only had one answer which isn't right as there are mean to be 2 answers.
I know the centre of the circle is (1,0)
I know the radius is 0.64
But the m is really making life harder.
Can anyone help me?
Pls...


Wait, isn't there infinite answers to m seeing as a tangent can be at any point on the circle? Unless you're given the values in the equation y=mx
Reply 9
Avatar for Pangol
Pangol
15
Original post by DennisChin
Wait, isn't there infinite answers to m seeing as a tangent can be at any point on the circle? Unless you're given the values in the equation y=mx

No - the given tangent equation passes through the origin, so a quick sketch shows that there are only two possibilities.
Original post by Pangol
No - the given tangent equation passes through the origin, so a quick sketch shows that there are only two possibilities.


Ah right yeah that makes sense
Original post by BlueAgent214
That's the problem. I don't know how to with the m in the way


An easier approach would be to substitute y = mx into the circle - that tells you where the line intersects. Then, because it is a tangent you can look at the discriminant and put that = 0. [two equal roots]

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