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# Core 1 Circle question

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1. Hi everyone! I'm stuck on this question:

The line is a tangent to the circle . Find the possible values of

We are not permitted to differentiate the equation of the circle to answer the question (that may be irrelevant anyway).

What I tried:
The line is a tangent, so there is only one solution to the pair of simultaneous equations of the line/circle. Hence, the final factorised form must be

Also, subbing in the equation of the line into the circle equation we get:

Then

But then I have no idea what to do, and I'm not even sure that my method is correct. Please could someone give hints, NOT FULL SOLUTIONS!!
2. (Original post by K-Man_PhysCheM)
Hi everyone! I'm stuck on this question:

The line is a tangent to the circle . Find the possible values of

We are not permitted to differentiate the equation of the circle to answer the question (that may be irrelevant anyway).

What I tried:
The line is a tangent, so there is only one solution to the pair of simultaneous equations of the line/circle. Hence, the final factorised form must be

Also, subbing in the equation of the line into the circle equation we get:

Then

But then I have no idea what to do, and I'm not even sure that my method is correct. Please could someone give hints, NOT FULL SOLUTIONS!!
Since the line is tangent to the circle, just plug in the equation of the line into the circle, get the form of and then the discriminant must be equal to 0 for equal roots - ie tangent.
3. (Original post by K-Man_PhysCheM)
*
form an expression for the discriminant of this, then equate it to zero... *
4. (Original post by the bear)
form an expression for the discriminant of this, then equate it to zero... *
Wow, can't believe I overlooked that...

Thank you, I can solve it now!
5. (Original post by RDKGames)
Since the line is tangent to the circle, just plug in the equation of the line into the circle, get the form of and then the discriminant must be equal to 0 for equal roots - ie tangent.
Thank you! Yeah, I had forgotten about the discriminant. Thank you again.
6. so after subbing the equation for Y back into the equation of the circle you should be able to got

x^2 + 9x^2 -6kx + k^2 - 10 = 0

which can now be written as
Ax^2 + Bx^2 + C = 0
10x^2 - (6k)x + (k^2 - 10) = 0

For this to be true the discriminant must be greater than or equal to zero

so B^2 - 4AC >/= 0

This should help find your values of k
7. (Original post by ljf98)
so after subbing the equation for Y back into the equation of the circle you should be able to got

x^2 + 9x^2 -6kx + k^2 - 10 = 0

which can now be written as
Ax^2 + Bx^2 + C = 0
10x^2 - (6k)x + (k^2 - 10) = 0

For this to be true the discriminant must be greater than or equal to zero

so B^2 - 4AC >/= 0

This should help find your values of k
they are specifically asking for the tangents so D = 0 not D ≥ 0
8. (Original post by ljf98)
so after subbing the equation for Y back into the equation of the circle you should be able to got

x^2 + 9x^2 -6kx + k^2 - 10 = 0

which can now be written as
Ax^2 + Bx^2 + C = 0
10x^2 - (6k)x + (k^2 - 10) = 0

For this to be true the discriminant must be greater than or equal to zero

so B^2 - 4AC >/= 0

This should help find your values of k
Thank you, though, as others have pointed out, it should the discriminant is equal to 0 (not greater than or equal to), because the line is tangent to the circle so there is only one solution. Thank you for the response, regardless.
9. Ah yes my bad, didnt see that
10. (Original post by K-Man_PhysCheM)
Wow, can't believe I overlooked that...

Thank you, I can solve it now!
OP what did you get for your answers? Tried this out myself and want to see if I got it
11. (Original post by surina16)
OP what did you get for your answers? Tried this out myself and want to see if I got it
You can just check them yourself by plugging it back into the quadratic and getting . Otherwise use a graphing software like Desmos.
12. (Original post by surina16)
OP what did you get for your answers? Tried this out myself and want to see if I got it
I got , and that's the answer in the back of the book too.

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