Anyone!!!

[davidgreen]

I been trying to do this question and it is causing me a lot of stress. Any1 have any ideas???

Circle C has centre (5,-1). the line L: 3x-4y+11=0 is a tangent to C.
(i) Show that the radius of C is 6.
(ii) The line x+py+1=0 is also a tangent to C.Find two possible values for p.

I can do part (i) its (ii) that I cant get to work out.

thanks to anyone who helps...

[Rassis]

Part (i) is easy to solve, but regarding part (ii), I think there is something wrong about the text. I tried hard and just couldn’t find a satisfactory answer. Actually I don’t think the general expression x+py+1=0 represents a line tangent to the circumference C as specified. Otherwise, what would the value of p be in the case of the line 3x-4y+11=0 (which undoubtedly is tangent to a circle with radius of 6 and centred in point 5;-1)? Would you mind to confirm?

Thanks,

Rui

[VBAhack]

Hi Rassis, I figured you would reply. You like math puzzles also!

I do believe there is a solution, but it isn't easy to obtain. In situations like this, it can be very helpful to graph the functions. By some trial and error I was able to get close by playing with P to convince myself that there are, indeed, solutions.

Attacking it from an algebraic standpoint, the approach I took was to find dy/dx of the circle and dy/dx of the line, then set them equal to each other, since the requirement is that the line must be tangent to the circle (i.e. equal slope). Thus, for the right x, y, and p, there are two solutions, since each x of the circle has 2 y's. Because we have equations relating x & y, we can reduce the above expression to x in terms of p, thus 2 variables.

There is another useful expression. That is, the x and y coordinates of the tangent point must lie on the line as well as the circle. Setting y(circle) = y (line), we get another expression of x in terms of p.

In the end we have 2 equations in 2 unkowns (x and p), but they are non-linear equations. Thus, no closed form solution. I believe that an iterative technique (Newton Raphson comes to mind) is needed.

[zaza]

Maybe I've missed the point here. But I don't think so.


Considering the line x + py + 1 =0 ... y = (-1/p) x - (1/p)

where does this line cross the x-axis? At y=0, regardless of the value of p. Hence all lines of this form pass through the point (-1,0).
We now know that the distance from here to the centre of the circle is sqrt(dx^2 + dy^2) = sqrt(37).

Draw radius lines from the centre of the circle to the intercepts of the two tangents. These have to have length 6, and hence the distance from the point (-1,0) to the intercept of the circle is sqrt(37-36) = 1.
Effectively, therefore, the tangent lines run from the point (-1,0) to the two intersections of the circles at (5,-1) radius 6 and at (-1,0) radius 1.

We can also now work out the angles between the line joining (-1,0) to the circle centre:

...and the tangent lines = arctan 6

...and the x axis = arctan(1/6)

The difference between these two gives the angle of elevation of the first tangent line from the x-axis. And we know the distance to intercept (1) so we can work out the coordinates of the intercept using cos and sin. From that we can work out the gradient, and hence calculate the value of p.

The second intercept is similar, but you add the two angles to give the declination between x-axis and tangent line....=90 degrees. Actually, this is obvious because the tangent line passes through (-1,0) and the centre of the circle is at (5,-1) with radius 6. Hence the leftmost point of the circle is at (-1,-1) and the tangent line points straight down.


But somebody else can actually work out the numbers...



zaza

[VBAhack]

Maybe I've missed the point here. But I don't think so

Correct you are. Straightforward geometry. The circle center, line pivot point (-1, 0) and tangent point to the circle form a right triangle. A second equal right triangle is formed with the 2nd tangent point. Thus:

P = tan(2*arctan(1/6)) = .343

The method I used was more complicated but had the same result:

p = tan(θ); 6 - sin(θ) - 6*cos(θ) = 0