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C2 Coordinate geometry question

Hi I'm struggling with this question and was hoping someone could give me an algebraic solution please?

A line, L, has the equation y=mx and a circle C has the equation x^+y^-6x-4y+9=0

given that L is a tangent to C find the possible values of M

So far I have got to making it (x-3)^+(y-2)^=4 centre- (3,2)
and have tried expanding it all out then substituting in mx for y and 3 for x, but I can't get the values the book gives in the answer section.

Thanks :smile:

(edited 10 years ago)

Reply 1

Jooooshy

I can't see a question.. There are just two equations!

EDIT: And the centre is (3,2)..

Reply 2

just george

What is the actual question? Where do the line and circle intersect?

Reply 3

Hufflepuff97

Original post by Jooooshy

I can't see a question.. There are just two equations!

EDIT: And the centre is (3,2)..

oh sorry I forgot to put it haha

Reply 4

username1258398

EDIT:
Nevermind; I assumed question was different.

If L, with gradient m, is a tangent to C, then it has only a single intersection with C. Thus, you can sub y = mx into the equation for C.

(x - 3)^2 + (mx - 2)^2 = 4

Simplifying gives:

(m^2 + 1)x^2 - (4m + 6)x + 9 = 0

For this equation to have only ONE solution, as is necessary since a tangent only touches a circle at ONE point, the discriminant must equal 0. Thus,

(4m + 6)^2 - 4(9)(m^2 + 1) = 0

You can simplify that to find a value for m.

(edited 10 years ago)

Reply 5

Hufflepuff97

sorry I put 2 instead of 4 for some reason in the centre and forgot to add the actual question but I've amended it all now :smile:

Reply 6

just george

A suggestion, although I havent tried it and it was a little while since i did C2 :tongue:

For the line L to be a tangent, it must touch the circle where a line from the point of intersection going through the centre of the circle has a gradient of -m, if that makes any sense?

So you can find the points where L intersects C, by substituting y=mx into the circle equation. You can then make an equation of another line that must go through the centre of the circle and have gradient -m. Could you then find the possible values of m by finding where both lines intersect the circle?

(edited 10 years ago)

Reply 7

Hufflepuff97

Sorry, I know what you mean in the first part of your answer but I don't really get the rest :s I expanded it all out to get

x^ -6x +9 +y^ -4y +4 = 4

then substituted in mx to get x^ -6x +9 + m^x^ -4mx =0 but now I don't know what to do :/

Reply 8

Hufflepuff97

Thank you strange banana :biggrin:

Reply 9

username1258398

Original post by Hufflepuff97

Thank you strange banana :biggrin:

No problem! ^^ By the way, where did you find that question? :redface:

Reply 10

chocolatemonkey7

is there more than 1 solution for m?
because i got m as 0 and m as 12/5.

(edited 10 years ago)

Reply 11

username1258398

Original post by chocolatemonkey7

is there more than 1 solution for m?
because i got m as 0 and m as 12/5.

I got those values as well; and if you think about it graphically, there must be two solutions for m, as any given point has to have two tangents going through it to a given circle.