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Circles and tangents - Help please!

I'm stuck on the following question:

The circle C has centre (8,-1) and passes through the point (4,1).

a) Find an equation for C.

b) Show that the line with equation x+2y+4=0 is a tangent to C.

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For a), I got (x-8)^2 + (y+1)^2 =20

However I have no idea how to answer b)!

Thanks!

Reply 1

insparato

Sub x = -2y - 4 into (x - 8)^2 + (y+1)^2 = 20, for a tangent to a circle there is only one repeated root ie b^2 - 4ac = 0

Reply 2

cuctacuctac

find the intersection(s) of the line with the circle. 1 point -> tangent

Reply 3

tba

Thanks for the help, but I'm still confused... sorry, I'm not very good at maths!

Reply 4

insparato

Okay, by substituting in x = -2y - 4(the tangent) into the circle equation, you are essentially trying to find where the circle intersects the tangent. You will get a quadratic, at the intersection point of a circle there can be only one point of intersection thus there can only be one solution.

The discriminant is a way of seeing whether a quadratic has 2 real, 1 repeated real root or no roots

b^2 - 4ac > 0 for 2 real roots
b^2 - 4ac = 0 - 1 repeated root (HENCE TANGENT)
b^2 - 4ac < 0 - No real roots

Reply 5

tba

I still don't get it... I know it's 'against the rules', but I would appreciate it if you could post a full solution so I can try and figure out the steps involved and how the final answer shows that the line is a tangent. Also is the answer for part a) correct in the first place?

Reply 6

insparato

Well, have you tried what i've suggested ? Do you understand why i'm telling you to substitute x = -2y - 4 into the circle equation ? Do you understand the significance of why im telling you to find b^2 - 4ac ?

Which bit are you not understanding? Try be a bit more specific to what you don't understand then i can help make you understand.

Reply 7

nota bene

Solve the quadratic.

Spoiler

No real roots means no intersection (hence not a tangent!)
Repeated root means the line cuts the circle at one point, i.e. is a tangent
Two real roots means the line cuts the circle twice, which isn't allowed if the line is to be a tangent.

edit: Sorry Aaron, didn't refresh!

Reply 8

tba

Well this is what I have done:

(-2y-4-8)^2+(y+1)^2=20

and ended up with:

y^2+10y +25 =0
(y+5)(y+5)=0

Does this show that the line is a tangent as both values for y are the same?

Reply 9

insparato

Indeed. There is only one repeated solution to a tangent intersecting a circle.

However i would've just done b^2 - 4ac = 0. Its easy if the quadratic doesn't factorise nicely or cant be factorised.

Reply 10

tba

Thanks for all the help everyone, (especially insparato)!

Reply 11

Nancarrow

insparato

However i would've just done b^2 - 4ac = 0. Its easy if the quadratic doesn't factorise nicely or cant be factorised.

True, but if there's only one repeated root, the quadratic has GOT to factorise nicely! :biggrin:

Reply 12

insparato

Nancarrow

True, but if there's only one repeated root, the quadratic has GOT to factorise nicely! :biggrin:

Ah, good point :p:

. Meh b^2 - 4ac to me just means plonking numbers in, whereas factorising requires a tad more thought.

I distinctly remember a question in FP2 quite similar to this but obviously harder. So i've just gone from there.

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