Math Is Fun Forum

Azalea

Guest

Proof of a Circle and Tangent

I need to show that the line y=mx+c touches the circle x²+y²=a² if c²=a²(1+m²).

What I believe is that the line will only touch the circle when it is at a tangent. Therefore, the point of intersection between of the two lines should mean that it will only cross at one point. Which we should be able to prove when we use the quadratic formula's discrminant b²-4ac. The only problem is I can't get the discrimant to equal 0.

       y=mx+c
x²+y²=a²

x²+m²x²+2cmx+c²=a²
x²+m²x²+2cmx+a²+a²m²=a²
(m²+1)x²+(2cm)x+a²m²=0

Therefore,
a=(m²+1)
b=2cm
c=a²m²

b²-4ac

4c²m²-[4(m²+1)(a²m²)]

4c²m²-(4m²a²+4m^4+4a²+4m²)

But how do you get rid of the a terms?

Please help and thank you in advance.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Proof of a Circle and Tangent

You've made a mistake in the third line.

x^2 + (mx+c)^2 = a^2
⇒ x^2 + m^2x^2 + 2mxc + c^2 = a^2
⇒ x^2(1+m^2) + x(2mc) + (c^2 - a^2) = 0

Azalea

Guest

Re: Proof of a Circle and Tangent

But, we are told c²=a²(1+m²)

So, for (c²-a²), then we would get: a²+a²m²-a² which is equal to a²m².

Which is what I got already...

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Proof of a Circle and Tangent

You multiplied out the 4ac term wrong and forgot to substitute the c portion of b^2:

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