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Circle geometry

Q3) "Show that the straight line y=x-8 is a tangent to the circle x^2+y^2-6x+2y=0 and find the coordinates of the point of contact."

First off, when plotted, the line is not actually a tangent but a chord of the circle, but I think that's just because they didn't bother being accurate for the sake of a question.
Next, here is what I have so far:
*gradient of line = 1
*m1m2=-1 so gradient of circle radius must be -1
*the circle equation becomes (x-3)^2-(y+1)^2 = 10, so the radius length is √(10).
*the centre of the circle is (3, -1)

I don't know where to go from here. I haven't actually done this topic since I joined Maths a month late, so all I know is what I read in the textbook today.

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Q4) "The straight line with equation y=kx is a tangent to the circle x^2+y^2-4x-4y+7=0. Find the possible values of k, giving your answers in the form a+b√7."

I haven't actually started this one yet, but I've looked at it and all I know is that I need to complete the square in the circle equation, which I'm sure I can do. I don't know what next, though. Apparently no one can do this one.

Should I try to work out the gradient of the radius and then use m1m2=1 to go from there?

Thanks!
Reply 1
Original post by JustJusty
Q3) "Show that the straight line y=x-8 is a tangent to the circle x^2+y^2-6x+2y=0 and find the coordinates of the point of contact."

First off, when plotted, the line is not actually a tangent but a chord of the circle, but I think that's just because they didn't bother being accurate for the sake of a question.

How are you meant to show that it's a tangent if it isn't a tangent!
Original post by JustJusty
Q3) "Show that the straight line y=x-8 is a tangent to the circle x^2+y^2-6x+2y=0 and find the coordinates of the point of contact."

First off, when plotted, the line is not actually a tangent but a chord of the circle, but I think that's just because they didn't bother being accurate for the sake of a question.
Next, here is what I have so far:
*gradient of line = 1
*m1m2=-1 so gradient of circle radius must be -1
*the circle equation becomes (x-3)^2-(y+1)^2 = 10, so the radius length is √(10).
*the centre of the circle is (3, -1)

I don't know where to go from here. I haven't actually done this topic since I joined Maths a month late, so all I know is what I read in the textbook today.

-----------------------------------------------------------------------------------------
Q4) "The straight line with equation y=kx is a tangent to the circle x^2+y^2-4x-4y+7=0. Find the possible values of k, giving your answers in the form a+b√7."

I haven't actually started this one yet, but I've looked at it and all I know is that I need to complete the square in the circle equation, which I'm sure I can do. I don't know what next, though. Apparently no one can do this one.

Should I try to work out the gradient of the radius and then use m1m2=1 to go from there?

Thanks!


The first question is actually wrong! Remember, a tangent only "meets" the circle in one place. This means that, if you substitute the eqn of the line in, then you should get a quadratic which has equal roots! However, that's not the case (I've checked).

For the second question, you just have to apply what I've said above to find k :smile:
Original post by Indeterminate
The first question is actually wrong! Remember, a tangent only "meets" the circle in one place. This means that, if you substitute the eqn of the line in, then you should get a quadratic which has equal roots! However, that's not the case (I've checked).

For the second question, you just have to apply what I've said above to find k :smile:


Yeah, my teacher said today to only work out the point(s) of contact for question 3 and skip the proving part, but it didn't really register what he was talking about until I tried to do the question. Is there a mathematical way (other than plotting a graph) to work out the point(s) of intersection though?

As for question 4, I substituted the line equation into the circle equation, but got
x^2 + k^2x^2 - 4x - 4kx + 7 = 0
so I'm not sure what to do with the k variable, especially the k^2x^2 part.
Reply 4
Original post by JustJusty
Yeah, my teacher said today to only work out the point(s) of contact for question 3 and skip the proving part, but it didn't really register what he was talking about until I tried to do the question. Is there a mathematical way (other than plotting a graph) to work out the point(s) of intersection though?

As for question 4, I substituted the line equation into the circle equation, but got
x^2 + k^2x^2 - 4x - 4kx + 7 = 0
so I'm not sure what to do with the k variable, especially the k^2x^2 part.

To find the intrsersection points, you need to solve the simultaneous equations

y=x-8
x^2+y^2-6x+2y=0

You can do this by substituting x-8 for y in the second equation.

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