# Thread: circle/line simultaneous equation

1. ## circle/line simultaneous equation

Sorry folks, I'm pretty new to this and I do wrack my brain before I ask the question. I've exhausted all I know with this one.

The circle c had equation

x^2+y^2=17

L is the line

x-4y-17=0

L is a tangent to C at the point T.

Find the coordinates of T.

I can see that the line L is going to form a sort of hypotenuse radius within the circle.

I'm really struggling here and I am trying very hard which makes it doubley frustrating. I've tried to go down the quadratic route by stating x in terms of y with the equation of the circle but it turns into 4y^2+136y+289=17 and that's just way over the top for me. I'm pretty sure it doesn't factorise either so I've ruled that out, I think?

Any help greatly appreciated. I'm way off the mark here for an exam on Thursday but it has never really been in my power to make this on my own, hard as I've tried.

2. ## Re: circle/line simultaneous equation

You were actually completely along the right lines! The only thing you did wrong was not trusting your instincts and having confidence in the method.

Here's the full explanation:

$\displaystyle C: x^2 + y^2 = 17$

$\displaystyle L: x - 4y - 17 = 0$

or more simply

$\displaystyle L: x = 4y + 17$

As line $\displaystyle L$ is a tangent to the circle $\displaystyle C$, we can work out the point they intersect by either using a simultaneous equation, or more simply, as you did, substituting $\displaystyle x$ in terms of $\displaystyle y$ into the equation for $\displaystyle C$.

hence,

$\displaystyle (4y + 17)^2 + y^2 = 17$

$\displaystyle 16y^2 + 136y + 289 + y^2 = 17$

now by tidying up,

$\displaystyle 17y^2 + 136y + 272 = 0$

all coefficients have a factor of 17, so we can simplify further:

$\displaystyle y^2 + 8y + 16 = 0$

n.b - I did not know off the top of my head that all those numbers are divisible by 17, but you can assume that it'll be something simple like that and then test to see if you are right (by quickly using your calculator). The questions in the books and exams are usually nice and tidy like that. So don't be put off when you see some quadratic with huge numbers - it's just trying to put you off!

Now this equation looks much more simple. We can easily solve to find $\displaystyle y$:

$\displaystyle (y+4)(y+4) = 0 \ or \ (y+4)^2 = 0$

We've got a repeated root which is great because we are told the line and circle on intersect at one point $\displaystyle T$ (the tangent).

so $\displaystyle y = -4$

substitution the value for $\displaystyle y$ into the equation for line $\displaystyle L$:

$\displaystyle x = 4(-4) + 17$

$\displaystyle x = -16 + 17 = 1$

So the coordinates of $\displaystyle T$ are $\displaystyle (-4,1)$.

3. ## Re: circle/line simultaneous equation

Thanks for that mate, that's really helpful. At the end though (excuse me if it's a silly question) but you say the coordinates are (-4,1). As x is 1 and y is -4 should the coordinates not be (1,-4)?

4. ## Re: circle/line simultaneous equation

Oh my, yes it should be haha. What an amateur mistake!

5. ## Re: circle/line simultaneous equation

Only an amateur would see it lol Thanks for the help mate, now I've got to find the other end of the diameter of point T. I really don't know how they expect us to acheive this in 3 weeks from a standing start. Frazzled lol

6. ## Re: circle/line simultaneous equation

The beuty of mathematics lies in one word ...simplicity..........

the circle has equation : x^2+y^2=17
the tangent to the circle at the point (x1,y1) has equation..xx1+yy1=17
comparing this equation with the given one x-4y=17 , we obtain x1 =1 y1=-4