Competition - post a photo you've taken of nature >>

I would love some help with this geometry problem!

Here is the question that I am stuck on:

Find the equation of the tangents drawn from the point (-4,3) to the circle x^2 + y^2 = 5

I know that the aim is to find the gradients of the tangents, thus enabling use of y - y1 = m(x - x1). However I have been struggling for a while to make any progress on this.

Thanks in advance for any help.

Scroll to see replies

Reply 1

lizard54142

Original post by Sm0key

Have you covered implicit differentiation? This would be the easiest way to get the gradient, or you could rearrange into the form y = ...

Reply 2

samb1234

Original post by Sm0key

Tangents meet a circle at right angles to the radius. Therefore you can draw a right angled triangle from the point to the centre of the circle with the tangent and them you can work out various things from that

Posted from TSR Mobile

Reply 3

Sm0key

Original post by lizard54142

Have you covered implicit differentiation? This would be the easiest way to get the gradient, or you could rearrange into the form y = ...

Yes I have. I'm afraid I don't follow how that would be of use here? I mean: what equation must I form to then implicitly differentiate?

Reply 4

XI Ki11JoY IX

Original post by Sm0key

Circle geometry is the best.

Well, first of all, the question has a few typos, the equation for a circle (with centre 0,0 is) x^2 + y^2 = r^2 , a quick check of the coordinates can also prove that r^2 is in fact 25.

Second, there is only 1 tangent at a point, so that might help.

right, y^2 = 25 - x^2

implicit differentiation gives 2y(dy/dx) = -2x

dy/dx = =2x/2y or -x/y or 4/3

y= 4/3x + k

k = -8

Tangent: y=4/3x -8

In from ax+bx+c=0

4x-3y-24=0

Edit: I'm dumb, sorry for being a jag, I'll insert another comment explaining what you are supposed to do.

(edited 8 years ago)

Reply 5

Sm0key

Original post by XI Ki11JoY IX

The question on the paper is definitely: "tangents from (-4,3) to the circle x^2 + y^2 = 5"
ie radius = sqrt(5).

Reply 6

username1162972

Original post by Sm0key

The question on the paper is definitely: "tangents from (-4,3) to the circle x^2 + y^2 = 5"
ie radius = sqrt(5).

There would be two tangents that would go this way.

Posted from TSR Mobile

Reply 7

username1732133

Any straight line through (-4,3) has equation

y-3=m(x+4)

y=mx+4m+3

.

You can substitute

mx+4m+3

for y in the equation of the circle to get a quadratic in x^2. Now think about the discriminant. This will allow you to find the two possible gradients.

There's probably an easier way but I'm still half asleep...

Spoiler

(edited 8 years ago)

How many marks?

Original post by XI Ki11JoY IX

I am pretty sure this is wrong.

Posted from TSR Mobile

Reply 10

username1732133

You can also express the equation of the tangent in the form

x \cos \theta + y \sin \theta = \sqrt{5}

Reply 11

lizard54142

Original post by Sm0key

Yes I have. I'm afraid I don't follow how that would be of use here? I mean: what equation must I form to then implicitly differentiate?

It would be of use here because you have an equation which is a "mixture" of x's and y's on one side, you don't have something nice like y = x^2 + 5. You don't have to form any equation, just jump straight in and start differentiating term by term.

Reply 12

sychosid

Original post by Sm0key

The question on the paper is definitely: "tangents from (-4,3) to the circle x^2 + y^2 = 5"
ie radius = sqrt(5).

you are wrong bro. the 2 equations are y=-2x-5 and 11y=-2x+25

Reply 13

XI Ki11JoY IX

Original post by Sm0key

The question on the paper is definitely: "tangents from (-4,3) to the circle x^2 + y^2 = 5"
ie radius = sqrt(5).

That's impossible.

Reply 14

username1162972

Original post by XI Ki11JoY IX

That's impossible.

No it isn't

Posted from TSR Mobile

Reply 15

XI Ki11JoY IX

Original post by physicsmaths

No it isn't

Posted from TSR Mobile

I'm un aware of any parametic equations for circles, I tried using my own variables for the coordinates and it got messy quickly, if I were to take a guess, there's content in his course that should help him easily, until then, I don't know.

Reply 16

Sm0key

Q9 circle & 2 tangents.pdf

It seems that some members don't trust my typing of the question - for everyone's benefit here is a scan of the question, with my sketch too.

Edit: After implicit diff. of the circle equation I get:

dy/dx = -x/y

But I can't see what 2 pairs of (x,y) values I need (obviously lying on the circumference) to be able get the two tangents' gradients?

(edited 8 years ago)

Reply 17

username1732133

Original post by Sm0key

See posts 8 and 11. Both hints I've given will get you to the answer quite quickly. Post again if you need more.

Reply 18

Sm0key

Original post by BuryMathsTutor

See posts 8 and 11. Both hints I've given will get you to the answer quite quickly. Post again if you need more.

Thanks for the pointers.

I've ended up with:

(1 + m^2)x^2 + (8m^2 + 6m)x + (16m^2 + 24m - 4) = 0

following the substitution. However I cannot decide what requirement I must place on the discriminant. I imagine that b^2 > 4ac is required to give 2 real solutions. However I can't work out how the above quadratic would handle 2 *tangents* as opposed to 2 lines with 2 intercepts each (ie 4 solutions). How do I know that consideration of the discriminant will produce only 2 solutions (corresponding to tangents)?

Reply 19

username1732133

Original post by Sm0key

If, for a given value of m, the quadratic has equal roots then this value produces a line which is a tangent to the circle. You need to solve discriminant = 0.

You have a small error in your equation by the way. The -4 should be +4.

You should get a quadratic in m: