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1. The circle with equation x2 + y2 – 12x – 10y + k = 0 meets the coordinate axes at exactly three points.
What is the value of k???
2. (Original post by Cameron45)
The circle with equation x2 + y2 – 12x – 10y + k = 0 meets the coordinate axes at exactly three points.
What is the value of k???
From that you can work out the centre of the circle.
Then, if it meets the coordinate axis exactly 3 times, this implies that it must meet once, and intersect twice. From the centre you should be able to tell the radius where it intersects twice and meets it once, you can use this to work out the value of k
Let me know if you get stuck!
3. Would the centre be (6,5)? Where do you go from there?
(Original post by Lemur14)
From that you can work out the centre of the circle.
Then, if it meets the coordinate axis exactly 3 times, this implies that it must meet once, and intersect twice. From the centre you should be able to tell the radius where it intersects twice and meets it once, you can use this to work out the value of k
Let me know if you get stuck!
4. because it meets the axes thrice, it follows that the radius must either equal the x coordinate of the centre or the y coordinate of the centre.
5. You can do it a different way - I think it's more about quadratic equation solving than circles.

Where it meets the axis, either x or y must be zero.

So, put x = 0, this will give you a quadratic in y (1)

Similarly, putting y = 0 gives you a quadriatic in x. (2)

Now, if there are only THREE places where it crosses the axes, then there are THREE solutions to those two quadratics which means one of them only has one solution. That means for one of them, the determinant is 0.

So, I took my two new quadratic equations, found the value of k which gave a determinant of zero.

That meant I had two possible values of k, one from (1) and one from (2)

I put the value of k from (1) back into (2) to try to solve for x
I put the value of (2) back into (1) to try to solve for y

You don't actually have to find them, just check the determinants again.

One of those two (not telling you which, try it) gave me a negative determinant which meant the equation would have NO solutions which means the original equation could not have THREE solutions.
The other one woried. I found the markscheme, checked and it's right.

Try it - if you can't get it to work out come back and ask again.
6. (Original post by Cameron45)
Would the centre be (6,5)? Where do you go from there?
Yes
So since they touch 3 times, and a circle is clearly very symmetrical then you know that in one direction it must intersect twice, and in the other direction it must just touch the axis. So what would the radius have to be if it is to touch one of the axis considering the centre? (there's two options here)
Then considering the two options, which one also means that it intersects the other axis?
7. Hi!
When I set x=0 and used b^2-4ac I got an answer of 36, and when I set y=0 I got 25. I don't understand what I do now.

(Original post by phys981)
You can do it a different way - I think it's more about quadratic equation solving than circles.

Where it meets the axis, either x or y must be zero.

So, put x = 0, this will give you a quadratic in y (1)

Similarly, putting y = 0 gives you a quadriatic in x. (2)

Now, if there are only THREE places where it crosses the axes, then there are THREE solutions to those two quadratics which means one of them only has one solution. That means for one of them, the determinant is 0.

So, I took my two new quadratic equations, found the value of k which gave a determinant of zero.

That meant I had two possible values of k, one from (1) and one from (2)

I put the value of k from (1) back into (2) to try to solve for x
I put the value of (2) back into (1) to try to solve for y

You don't actually have to find them, just check the determinants again.

One of those two (not telling you which, try it) gave me a negative determinant which meant the equation would have NO solutions which means the original equation could not have THREE solutions.
The other one woried. I found the markscheme, checked and it's right.

Try it - if you can't get it to work out come back and ask again.
8. Yes - you are right so far.

I said if k = 25, then the quadratic I got for x only has one solution so the one I got for y must give me two solutions which means the determinant part of that quadratic in y is positive. - check it by substutuing k=25 into it.

Similarly, if k=36 then the quadratic I got for x must give two solutions so has a positive determinant. Chck it by subs k=36.

One of them gives you a negative determinant so no solution, therefore the other one must be correct.
9. So, do I put the value of k into y^2-10y+k and x^2-12x+k?

(Original post by phys981)
Yes - you are right so far.

I said if k = 25, then the quadratic I got for x only has one solution so the one I got for y must give me two solutions which means the determinant part of that quadratic in y is positive. - check it by substutuing k=25 into it.

Similarly, if k=36 then the quadratic I got for x must give two solutions so has a positive determinant. Chck it by subs k=36.

One of them gives you a negative determinant so no solution, therefore the other one must be correct.
10. (Original post by Cameron45)
So, do I put the value of k into y^2-10y+k and x^2-12x+k?

Yes, put the 36 into the y² one, then put the 25 into the x² one and you should find that one of them gives a positive determinant, so is correct, and the other gives a negative determinant.
11. Yeah, the 36 into y^2 gives you a negative, so 25 must be right because it gives you 3 solutions?

(Original post by phys981)
Yes, put the 36 into the y² one, then put the 25 into the x² one and you should find that one of them gives a positive determinant, so is correct, and the other gives a negative determinant.
12. (Original post by Cameron45)
Yeah, the 36 into y^2 gives you a negative, so 25 must be right because it gives you 3 solutions?
Yes!!!

here is the paper it came from https://www.sqa.org.uk/files_ccc/H_M...ellcheckon.pdf Q18 I think

and here the MS https://www.sqa.org.uk/pastpapers/pa...cs_mi_2015.pdf

just so you know I'm not kidding.
13. Same paper i'm working from just now!

(Original post by phys981)
Yes!!!

here is the paper it came from https://www.sqa.org.uk/files_ccc/H_M...ellcheckon.pdf Q18 I think

and here the MS https://www.sqa.org.uk/pastpapers/pa...cs_mi_2015.pdf

just so you know I'm not kidding.
14. (Original post by Cameron45)
Same paper i'm working from just now!
You're welcome

But hide that markscheme and do it properly
15. Will do haha!

(Original post by phys981)
You're welcome

But hide that markscheme and do it properly
16. Lemur14's method is the nicest and fastest for this question (a diagram will help). There's no need to consider discriminants.

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