Equation of a tangent to a circle

paulb203 · 2024-05-11 05:05:43

The line L is a tangent to the circle
(X^2) + (y^2) = 68 at the pont P
P is the point (2,8)
Work out the equation of the line L

Q. How do we do this without knowing where the centre of the circle is?

Bob · 2024-05-11 06:13:29

From the format the centre is (0,0). The general equation is (x-a)^2 + (y-b)^2 = r^2. Centre (a,b)

Bob

mathxyz · 2024-05-11 08:16:44

paulb203 wrote:

The line L is a tangent to the circle
(X^2) + (y^2) = 68 at the pont P
P is the point (2,8)
Work out the equation of the line L
Q. How do we do this without knowing where the centre of the circle is?

Is this precalculus or calculus l?

paulb203 · 2024-05-13 10:44:12

@mathxyz

Pre-calculus. GCSE Higher Tier (UK)

paulb203 · 2024-05-13 10:46:33

Bob wrote:

From the format the centre is (0,0). The general equation is (x-a)^2 + (y-b)^2 = r^2. Centre (a,b)
Bob

Thanks, Bob.
I get that the a and b in the general equation are the centre of the circle, but how do we know a=0, and b=0?

Bob · 2024-05-13 18:05:16

(x+a)^2= x^2 + 2ax + a^2

The 2ax term is missing . Only explanation is a=0. Similarly for 2by.

Bob

mathxyz · 2024-05-14 01:32:23

Bob wrote:

(x+a)^2= x^2 + 2ax + a^2
The 2ax term is missing . Only explanation is a=0. Similarly for 2by.
Bob

Very cool. Interesting question.

1. More conditional probability today.

2. Tomorrow, maybe, back to my college algebra textbook.

paulb203 · 2024-05-14 22:25:05

Bob wrote:

(x+a)^2= x^2 + 2ax + a^2
The 2ax term is missing . Only explanation is a=0. Similarly for 2by.
Bob

Thanks. Where does (x+a) come from (in the general equation it's x-a)?

Bob · 2024-05-14 23:32:49

Yes, sorry I should have said (x-a)^2. But the no "-2xa" still applies.

Bob

paulb203 · 2024-05-16 05:53:16

Bob wrote:

Yes, sorry I should have said (x-a)^2. But the no "-2xa" still applies.
Bob

Thanks, Bob.

What does, "The -2xa term is missing" mean?

We expanded (x-a)^2 and got x^2-ax-ax+a^2 which simplifies to x^2-2ax+a^2

And then..?

Bob · 2024-05-16 20:25:15

hi paulb203

Now compare that with

If these represent the same circle then by subtracting

If this was just a random equation then there are lots of values for x, y, a and b that would work.

But it's for points lying on a circle so we don't have that much freedom.

Because x and y can be lots of values but everything else if fixed, the only way that can work is if a=b=0 thus making the x and y terms disappear.** That means the original equation has r^2 = 68 and centre (0,0).

I've never seen a formal proof for the ** statement; just used it loads of times. I now realise that's not fully satisfactory so I'm going to work on filling in the proper details of a proof. (It's accepted in GCSE without proof) Meanwhile you can use what I've told you to do the problem.

LATER EDIT:

Here's the proof:

If you have {a circle, radius 68 centred on the origin} = statement A
then you can use Pythag to get x^2 + y^2 = 68 {statement B} as the equation.

So A => B

Our problem is does B => A

So we have an equation x^2 + y^2 = 68. Let's say (p,q) is a point satisfying this equation. ie p^2 + q^2 = 68.

As a negative squared gives the same result as the same number but positive squared this means that

(p,-q) (-p,q) and (-p, -q) all also fit the equation.

Join (p,-q) to -p,q) The midpoint of this line is ( (p-p)/2 , (q-q)/2 ) = (0,0). Similarly (0,0) is the midpoint of the line joining (-p,-q) to (p,q).

The distance from (0,0) to each of these points is the same. So all four lie on a circle centred on the origin. But this is true for all p and q fitting the equation. Thus the equation is for a circle, radius 68 centred on (0,0)

Bob

paulb203 · 2024-05-17 06:39:53

Great answer. Thanks, Bob.

I think I'm getting there, understanding the first part. I'll need to come back for another look. And another

Last edited by paulb203 (2024-05-17 06:40:19)

Bob · 2024-05-17 07:31:17

Bob

mathxyz · 2024-05-18 10:19:57

Bob wrote:

Bob

Isn't the equation of a tangent line to a circle a calculus 1 topic?

Bob · 2024-05-18 19:25:41

This question doesn't need calculus. It does make use of the rule that if m1 times m2 = -1 for gradients of two lines, they are perpendicular.

Bob

mathxyz · 2024-05-19 01:50:07

Bob wrote:

This question doesn't need calculus. It does make use of the rule that if m1 times m2 = -1 for gradients of two lines, they are perpendicular.
Bob

Ok. Copy. Nice problem. Interestingly challenging. I will post a few questions later today. Gotta go to the supermarket first.

Last edited by mathxyz (2024-05-19 01:50:23)

Math Is Fun Forum

#1 2024-05-11 05:05:43

Equation of a tangent to a circle

#2 2024-05-11 06:13:29

Re: Equation of a tangent to a circle

#3 2024-05-11 08:16:44

Re: Equation of a tangent to a circle

#4 2024-05-13 10:44:12

Re: Equation of a tangent to a circle

#5 2024-05-13 10:46:33

Re: Equation of a tangent to a circle

#6 2024-05-13 18:05:16

Re: Equation of a tangent to a circle

#7 2024-05-14 01:32:23

Re: Equation of a tangent to a circle

#8 2024-05-14 22:25:05

Re: Equation of a tangent to a circle

#9 2024-05-14 23:32:49

Re: Equation of a tangent to a circle

#10 2024-05-16 05:53:16

Re: Equation of a tangent to a circle

#11 2024-05-16 20:25:15

Re: Equation of a tangent to a circle

#12 2024-05-17 06:39:53

Re: Equation of a tangent to a circle

#13 2024-05-17 07:31:17

Re: Equation of a tangent to a circle

#14 2024-05-18 10:19:57

Re: Equation of a tangent to a circle

#15 2024-05-18 19:25:41

Re: Equation of a tangent to a circle

#16 2024-05-19 01:50:07

Re: Equation of a tangent to a circle

Board footer