Tangents to Circles: Find the length of chord AB

[Monkeyseat]

The line y = mx is a tangent to the circle x^2 + y^2 - 10y + 16 = 0

with m = +/- (4/3)

The tangents meet the circle at points A and B. Find the length of AB.

Sorry I can't show working because I don't get it - what does it mean AB? I know about conditions for a line to meet a circle and the length of the tangents from a point to a circle vaguely but I don't understand this. Go slowly please lol. :mrgreen:

Many thanks for any help.

 

[Denis]

Re: Tangents to Circles

Why don't you "experiment":
get a sheet of graph paper
pick a radius and draw a circle (stick to top right quadrant: easier)
pick a point P outside circle (known coordinates)
draw the 2 tangents from point P : label tangent points A and B

Now you know PA = PB; and you "see" that AB is simply a chord;
and you "see" that PAB is an isosceles triangle; and ... and ... and ...

 

[Monkeyseat]

Hi, I'm still a bit stuck.

I re-arranged the formula and got:

x^2 + (y-5)^2 = 9

So the radius is 3 and the centre is (0,5).

Now I drew the circle and noticed the triangle, but I'm stuck where to go next - I only have the length of one side (3 - the radius) and don't know the co-ordinates to figure the others out.

Any more suggestions? By the way - should the diagram look a bit like this?

Where the tangents are y = -(4/3)x and y = (4/3)x and the blue line is the chord.

Do I need to find A and B by factorising? It all went wrong when I tried that. :mrgreen: Or find an angle?

I'm nearly there. Any help is appreciated - it's due tomorrow so I'm trying to get it done.

 

[galactus]

You found the circle equation in terms of x. Good.

Now, we're dealing with the bottom half of the circle.

Solve the equation \(\displaystyle \frac{4}{3}x=-\sqrt{9-x^{2}}+5\) for x.

The other point will be the negative of that. You can then easily find y by plugging x back into the equation. Once you have the coordinates, use the distance formula to find the distance between the points and, therefore, the chord length.

 

[Monkeyseat]

You found the circle equation in terms of x. Good.

Now, we're dealing with the bottom half of the circle.

Solve the equation \(\displaystyle \frac{4}{3}x=-\sqrt{9-x^{2}}+5\) for x.

The other point will be the negative of that. You can then easily find y by plugging x back into the equation. Once you have the coordinates, use the distance formula to find the distance between the points and, therefore, the chord length.

I'm not sure where you got that equation from or how to solve it because there are 2 co-efficients of x and one is a wierd value (4/3)... Sorry.

What happens when you \(\displaystyle -\sqrt{-x^{2}}\)? Is it just x? Or -x? :? So when square rooted the right side would give -3 + x + 5?

I've gone wrong somewhere, I got the distance as 20 which is wrong. Could you possibly explain it a bit more? Thanks. I don't know where the formula comes from and I am not sure on solving it. Would the x value be substituted into the tangent or circle equation?

 

[stapel]

I'm not sure where you got that equation from...

What is the equation for the bottom half of the given circle? What is the equation for the tangent lines, given the slope and that the tangents pass through the origin? :?:

How does one find the intersection point(s) of two curves? :wink:

I'm afraid I don't understand your other questions...?

Eliz.

 

[Monkeyseat]

I'm not sure where you got that equation from...

What is the equation for the bottom half of the given circle? What is the equation for the tangent lines, given the slope and that the tangents pass through the origin? :?:

How does one find the intersection point(s) of two curves? :wink:

I'm afraid I don't understand your other questions...?

Eliz.

I'm lost here now... I know the equation of the tangent lines: y = +-(4/3)x. I don't know the equation for the bottom half of a circle. The way we've been doing these at the moment is substitution, i.e. I did this:

x^2 + (y-5)^2 = 9

I replaced y with 4/3x and then when I tried to expand it there were a lot of fractions and everything went wrong. I ended up with:

(25/9)x^2 - (40/3)x + 25 = 9

I have no idea how to solve that mess.

I didn't know how to solve galactus' equation:

\(\displaystyle \frac{4}{3}x=-\sqrt{9-x^{2}}+5\) for x.

I didn't know how he got it and what would happen when you do \(\displaystyle -\sqrt{-x^{2}}\). I also don't know which equation you are meant to put the x value in when you get it (if I ever do) - the circle or the tangent?

 

[Denis]

I didn't know how to solve galactus' equation:
\(\displaystyle \frac{4}{3}x=-\sqrt{9-x^{2}}+5\) for x.
I didn't know how he got it and what would happen when you do \(\displaystyle -\sqrt{-x^{2}}\). I also don't know which equation you are meant to put the x value in when you get it (if I ever do) - the circle or the tangent?

Well, if you can't solve this:
4x / 3 = -sqrt(9 - x^2) + 5 , then you're in trouble and not ready for such a problem!

Multiply each side by 3:
4x = -3sqrt(9 - x^2) + 15
subtract 15 from each side:
4x - 15 = -3sqrt(9 - x^2)
square both sides:
16x^2 - 120x + 225 = 9(9 - x^2)
16x^2 - 120x + 225 = 81 - 9x^2

Are you ok now ?

 

[Monkeyseat]

I didn't know how to solve galactus' equation:
\(\displaystyle \frac{4}{3}x=-\sqrt{9-x^{2}}+5\) for x.
I didn't know how he got it and what would happen when you do \(\displaystyle -\sqrt{-x^{2}}\). I also don't know which equation you are meant to put the x value in when you get it (if I ever do) - the circle or the tangent?

Well, if you can't solve this:
4x / 3 = -sqrt(9 - x^2) + 5 , then you're in trouble and not ready for such a problem!

Multiply each side by 3:
4x = -3sqrt(9 - x^2) + 15
subtract 15 from each side:
4x - 15 = -3sqrt(9 - x^2)
square both sides:
16x^2 - 120x + 225 = 9(9 - x^2)
16x^2 - 120x + 225 = 81 - 9x^2

Are you ok now ?

I get how you expanded it now, just had a forgetful moment. Okay so that gives the values x = 2.4 and y = 3.2. How do I find the other values now? Use (-4/3) instead? I thought that would give the same value when squared (16x^2).

I still don't know where the formula originated from though... :? How? I thought AB was tangent length at first - or that's what it sounded like. The answer is 4.8 apparently.

This is probably pointless but I re-arranged y = (4/3)x to x = (3/4)y and input it in x^2 + (y-5)^2 = 9 as (3/4y)^2 + (y-5)^2 = 9. I got the same answer on the calculator after expanding and using the quadratic formula - we're not meant to do it on them but I was just checking, some of these questions in this book are a bit daft sometimes.

If anyone could just clear this up it would be greatly appreciated - it's due in tomorrow morning.

 

[galactus]

Where the formula originated from?. You had it, I just solved for y. Plug your x back in to find y.

 

[Monkeyseat]

Where the formula originated from?. You had it, I just solved for y. Plug your x back in to find y.

I don't know where you got it from - it just lost me... Can you please show me? When you say "the equation for the bottom half of the given circle" I don't remember doing that. You solved it for y? Did I have it for x? :?

I solved it using the one you showed to get y = 3.2 and x = 2.3 but how do I get the others?

Sorry to ask so many questions. If you/anyone could go through it and clear it up that would be great.

Thanks again for being patient.