Length of Tangents to Circles: root of h^2+k^2+2hf+2gk+c

[Monkeyseat]

Sorry for posting another question so soon, but out of all of them there was only 2 I couldn't do and they're due in quite soon so I thought you guys might be able to help.

I scanned it because it might be a bit unclear if I typed it.

I'm not sure what to do, I tried completing the square to get it as (x-q) + (y-r) = radius but it failed. Sorry for little working - I don't know where to start.

Any help is appreciated. Go slowly if possible. Thanks.

 

[stapel]

I scanned it because it might be a bit unclear if I typed it.

Not at all! Using the formatting explained here or here, you can express the exercise quite clearly:

. . . . .Show that the lengths of the tangents from the point (h, k)
. . . . .to the circle x^2 + y^2 + 2fx + 2gy + c = 0 are given by:

. . . . .sqrt[h^2 + k^2 + 2fh + 2gk + c]

I'm not sure what to do, I tried completing the square to get it as (x-q) + (y-r) = radius but it failed...

Actually, you were on the right track! :wink:

After you completed the square, you should have had the following equation:

. . . . .(x + f)[sup:1k5ka7mw]2[/sup:1k5ka7mw] + (y + g)[sup:1k5ka7mw]2[/sup:1k5ka7mw] = f[sup:1k5ka7mw]2[/sup:1k5ka7mw] + g[sup:1k5ka7mw]2[/sup:1k5ka7mw] - c

So the center is at (-f, -g), and the radius is given by r[sup:1k5ka7mw]2[/sup:1k5ka7mw] = f[sup:1k5ka7mw]2[/sup:1k5ka7mw] + g[sup:1k5ka7mw]2[/sup:1k5ka7mw] - c.

Now draw the picture. You've got a circle centered at point (-f, -g). Label some point on its circumference as (x, y), and label some point off to the side as (h, k). Draw the radius line from the circle's center to (x, y), and the tangent line (remembering that it's at right angles to the circle) through (h, k) and (x, y).

You should have a right triangle with vertices (-f, -g), (x, y), and (h, k).

You have an expression for the radius (the side of the triangle between (-f, -g) and (x, y) on the circumference). Use the Distance Formula (squared) to get an expression, in terms of f, g, h, and k, for the square of the length of the hypotenuse.

Plug the squared-radius expression and the squared-hypotenuse expression into the Pythagorean Theorem, and solve for the expression which gives the length of the third side. After simplification, you should get the required answer!

Eliz.

P.S. No, this was not obvious! I made a few missteps in figuring this out, too!

 

[Monkeyseat]

Re:

After you completed the square, you should have had the following equation:

. . . . .(x + f)[sup:2jqh0ihi]2[/sup:2jqh0ihi] + (y + g)[sup:2jqh0ihi]2[/sup:2jqh0ihi] = f[sup:2jqh0ihi]2[/sup:2jqh0ihi] + g[sup:2jqh0ihi]2[/sup:2jqh0ihi] - c

So the center is at (-f, -g), and the radius is given by r[sup:2jqh0ihi]2[/sup:2jqh0ihi] = f[sup:2jqh0ihi]2[/sup:2jqh0ihi] + g[sup:2jqh0ihi]2[/sup:2jqh0ihi] - c.

Now draw the picture. You've got a circle centered at point (-f, -g). Label some point on its circumference as (x, y), and label some point off to the side as (h, k). Draw the radius line from the circle's center to (x, y), and the tangent line (remembering that it's at right angles to the circle) through (h, k) and (x, y).

You should have a right triangle with vertices (-f, -g), (x, y), and (h, k).

You have an expression for the radius (the side of the triangle between (-f, -g) and (x, y) on the circumference). Use the Distance Formula (squared) to get an expression, in terms of f, g, h, and k, for the square of the length of the hypotenuse.

Plug the squared-radius expression and the squared-hypotenuse expression into the Pythagorean Theorem, and solve for the expression which gives the length of the third side. After simplification, you should get the required answer!

Many thanks for the in depth reply. I got that equation for the cirlce, I just thought it was wrong at first lol! :mrgreen: Drawing the diagram really helps to visualise it.

Anyway, for the distance from (h, k) to (-f, -g) I got:

sqrt.(f^2 + h^2 +2fh + g^2 + k^2 + 2gk)

One thing I was wondering, do I need to figure out the distance (x, y) to (-f, -g) - I realised this was the radius when I drew the diagram:

sqrt.(f^2 + x^2 + 2fx +g^2 + y^2 +2gy)

I think this is irrelevant because we already have the radius, but why does this radius differ to f^2 + g^2 - c^2?

Anyway, then I did Pythag:

sqrt.(f^2 + h^2 +2fh + g^2 + k^2 + 2gk) - f^2 - g^2 + c

By the way, is the "- f^2 - g^2 + c" correct? I thought it was "f^2 + g^2 + c" (i.e. a^2 + b^2 = c^2) because we were finding the hypotenuse and the radius and the other line were a and b? Or is it "- f^2 - g^2 - c".

Anyway this cancelled down to:

sqrt.(h^2 + k^2 + 2fh + 2gk + c)

Would we get rid of the squared sign on h and k and just put the rest in the square root i.e. h + k + sqrt.(2fh + 2gk + c)? Although that is not how the question puts it. So I got the right answer, many thanks! Just a few points I am unclear on, if you could help that would be great. I put it in underline because usually my answers are quite hard to follow lol!

Once again, thanks. Any more assistance to clear up the last bits is appreciated. All typos corrected hopefully.

 

[stapel]

...do I need to figure out the distance (x, y) to (-f, -g)

Yes: You need two sides of the triangle, in order to find the third.

I realised this was the radius when I drew the diagram:

sqrt.(f^2 + x^2 + 2fx +g^2 + y^2 +2gy)

I think this is irrelevant because we already have the radius, but why does this radius differ to f^2 + g^2 - c^2?

Because the expression was found in a different way, using different information. If we had numerical values for all the numbers, the expressions would necessarily evaluate to the same value.

Note: "f[sup:2ghocz6o]2[/sup:2ghocz6o] + g[sup:2ghocz6o]2[/sup:2ghocz6o] - c[sup:2ghocz6o]2[/sup:2ghocz6o]" is not (quite) the value of the square of the radius.

Anyway, then I did Pythag:

sqrt.(f^2 + h^2 +2fh + g^2 + k^2 + 2gk) - f^2 - g^2 + c

You will need to use the expression for the square of the hypotenuse!

Eliz.

 

[Monkeyseat]

Re:

...do I need to figure out the distance (x, y) to (-f, -g)

Yes: You need two sides of the triangle, in order to find the third.

But that would just give (f^2 + x^2 + 2fx +g^2 + y^2 +2gy)? Is that not just sqrt.(f^2 + g^2 - c^2)?

Note: "f[supg2ys4n2]2[/supg2ys4n2] + g[supg2ys4n2]2[/supg2ys4n2] - c[supg2ys4n2]2[/supg2ys4n2]" is not (quite) the value of the square of the radius.

Is it just sqrt.(f^2 + g^2 - c^2)?

sqrt.(f^2 + h^2 +2fh + g^2 + k^2 + 2gk) - f^2 - g^2 + c
You will need to use the expression for the square of the hypotenuse!

So my answer was wrong? Can you show me please?

When I used it for the sqare of the hypotenuse it didn't work:

hyp = sqrt.(h^2 + f^2 + 2fh + g^2 + k^2 +2gk) + sqrt.(f^2 + g^2 - c^2)

And that doesn't cancel to the answer I had before...

Could you please show me what to do now? I'm just not sure. I've drawn a diagram but can't post it atm. Thanks.

Btw, could the distance of a line between 2 points formula be used?

 

[stapel]

Re: Re:

You need two sides of the triangle, in order to find the third.

But that would just give (f^2 + x^2 + 2fx +g^2 + y^2 +2gy)? Is that not just sqrt.(f^2 + g^2 - c^2)?

How are you getting that these two expressions are equivalent? And where does "f[sup:30fumumi]2[/sup:30fumumi] + g[sup:30fumumi]2[/sup:30fumumi] - c[sup:30fumumi]2[/sup:30fumumi]" keep coming from? Please show your work.

Note: "f[sup:30fumumi]2[/sup:30fumumi] + g[sup:30fumumi]2[/sup:30fumumi] - c[sup:30fumumi]2[/sup:30fumumi]" is not (quite) the value of the square of the radius.

Is [the square of the radius] just sqrt.(f^2 + g^2 - c^2)?

No. The square of the radius will not involve any square roots; it will involve only the expression I gave you earlier.

So my answer was wrong? Can you show me please?

...Btw, could the distance of a line between 2 points formula be used?

I've given you the corrected completed-square form, the expression for the square of the radius, and step-by-step instructions. Since you don't appear to be aware of the squared-radius expression, or that I plainly specified that you'll need to use the Distance Formula, it would appear that you have not yet read or followed the explanation provided earlier.

I'm afraid I'm not one of those people who "does" students' assignments for them. I'll be glad to try to provide more advice, but it would be good if you showed that you'd tried to use at least some of what you've already been given. :wink:

Thank you for your understanding and cooperation.

Eliz.

 

[Monkeyseat]

You need two sides of the triangle, in order to find the third.

But that would just give (f^2 + x^2 + 2fx +g^2 + y^2 +2gy)? Is that not just sqrt.(f^2 + g^2 - c^2)?

How are you getting that these two expressions are equivalent? And where does "f[sup3cvyd99]2[/sup3cvyd99] + g[sup3cvyd99]2[/sup3cvyd99] - c[sup3cvyd99]2[/sup3cvyd99]" keep coming from? Please show your work.

Using the distance formula to figure the distance between (x, y) to (-f, -g) I got:

sqrt.(f^2 + x^2 + 2fx +g^2 + y^2 +2gy) but I thought that was the same as the radius sqrt.(f^2 + g^2 - c) like you said - "Because the expression was found in a different way, using different information. If we had numerical values for all the numbers, the expressions would necessarily evaluate to the same value." I was told that sqrt.(f^2 + g^2 - c) was the radius...? I don't understand why r^2 gives the radius. I know that you square it for pythag. I meant that not sqrt.(f^2 + g^2 - c^2) before, sorry.

I don't expect you to do it at all and I read your reply numerous times. I knew I had to use the distance formula to get the length of one side, but I didn't know whether it had to be used again!

-------

I think I know the problem, look at this diagram:

You are referring to the distance between (-f, -g) and (h, k) as the hypotenuse, yet when I showed you what I had done I thought we were finding the hypotenuse. Why is this not the case? If my diagram is wrong and the normal meets the tangent at 90 degrees then it should be:

Where x is the missing side...

x^2 = (f^2 + h^2 + 2fh + g^2 + k^2 + 2gk) - (f^2 + g^2 -c)
x^2 = h^2 + 2fh + k^2 + 2gk + c

By the way, is it +c because it is take a negative i.e. the 2 whole brackets taking away so (...2gk) - (....-c) or is it just (2gk - f^2)? I think it's plus c because it's like multiplying the brackets by -1. I'm just checking incase I did the wrong method and because my teacher wrote the final answer as sqrt. (h^2 + k^2 + 2fh + 2gk - c) however I think that was just a mistake. Would it be better to leave one side as (h + f)^2 + (k+g)^2 or multiply it out to f^2 + h^2 + 2fh + g^2 + k^2 + 2gk as I did?

Therefore:

x = sqrt. (h^2 + k^2 + 2fh + 2gk + c)

Correct?

Can I just clear up, is the radius sqrt.(f^2 + g^2 - c)? And does the distance between (x, y) to (-f, -g) need to be calculated using the distance formula or can the radius be used (but squared?)?

Thank you. Sat here for over an hour typing this up lol... Reply appreciated. I completely forgot about the normal/tangent thing - sorry! Sorry for my slow reply and asking so many questions. I was kind of confused by the end of all this lol, not really about the answer, just because there were so many questions flying around lol. And all the typos in my previous replies... :? I've proof-read this one a few times. Let me know if I've missed anything.

EDIT:

Sorry, should be (-f, -g) not (-f, g).

 

[stapel]

You might want to review the definition of a tangent line: As was mentioned in the first reply you received, the tangent is at right angles to the curve to which it is tangent. So where should your right-angle marker be in your drawing? Which side is actually the hypotenuse?

Eliz.

 

[Monkeyseat]

Re:

You might want to review the definition of a tangent line: As was mentioned in the first reply you received, the tangent is at right angles to the curve to which it is tangent. So where should your right-angle marker be in your drawing? Which side is actually the hypotenuse?

Eliz.

I realised the correct definition and that my diagram was wrong as I mentioned but I don't have time to draw it again so I've done a bit of a quick edit above, the right angle should be where the normal from the centre meets the tangent to the circle. The hypotenuse is the distance between (-f, -g) and (h, k) or f^2 + h^2 + 2fh + g^2 + k^2 + 2gk by using the distance formula, so we are not trying to find the hypotenuse, it is one of the other sides so it is c^2 (hyp) - b^2 (radius) = a^2 (the side we are looking for). You did mention it was a right angle in the first reply but I forgot it was where the normal and tangent met. Is the rest of my working correct in the previous reply (bit more detail there but I'll copy it for here)?

x^2 = (f^2 + h^2 + 2fh + g^2 + k^2 + 2gk) - (f^2 + g^2 -c)
x^2 = h^2 + 2fh + k^2 + 2gk + c
x = sqrt. (h^2 + k^2 + 2fh + 2gk + c)

Much appreciated if you could confirm and help me with my other queries. Sorry for asking so many questions, I just want to understand it. I'm not trying to copy you or anything, this topic was finished ages ago at school so it's not due in or anything, I'm just wondering that's all and would just like it if you could check whether I have I got it right. Thanks. Let me know if there's any typos. I keep spotting them no matter how careful I check heh, should be alright. Sorry about that.

 

[Deleted member 4993]

Re: Re:

You might want to review the definition of a tangent line: As was mentioned in the first reply you received, the tangent is at right angles to the curve to which it is tangent. So where should your right-angle marker be in your drawing? Which side is actually the hypotenuse?

Eliz.

I realised the correct definition and that my diagram was wrong as I mentioned but I don't have time to draw it again so I've done a bit of a quick edit above, the right angle should be where the normal from the centre meets the tangent to the circle. The hypotenuse is the distance between (-f, -g) and (h, k) or f^2 + h^2 + 2fh + g^2 + k^2 + 2gk by using the distance formula, so we are not trying to find the hypotenuse, it is one of the other sides so it is c^2 (hyp) - b^2 (radius) = a^2 (the side we are looking for). You did mention it was a right angle in the first reply but I forgot it was where the normal and tangent met. Is the rest of my working correct in the previous reply (bit more detail there but I'll copy it for here)?

x^2 = (f^2 + h^2 + 2fh + g^2 + k^2 + 2gk) - (f^2 + g^2 -c)
x^2 = h^2 + 2fh + k^2 + 2gk + c
x = sqrt. (h^2 + k^2 + 2fh + 2gk + c)

Much appreciated if you could confirm and help me with my other queries. Sorry for asking so many questions, I just want to understand it. I'm not trying to copy you or anything, this topic was finished ages ago at school so it's not due in or anything, I'm just wondering that's all and would just like it if you could check whether I have I got it right. Thanks. Let me know if there's any typos. I keep spotting them no matter how careful I check heh, should be alright. Sorry about that.

Looking good.....

What were your "other queries"?