SOLUTION: Dear Genii
I have what I fear is a rather silly question. The other day I came across this question on a GCSE maths paper:
x-y = k
x^2+y^2-9=0
These simultaneous equations h
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Quadratic Equations and Parabolas
-> SOLUTION: Dear Genii
I have what I fear is a rather silly question. The other day I came across this question on a GCSE maths paper:
x-y = k
x^2+y^2-9=0
These simultaneous equations h
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Question 1208761: Dear Genii
I have what I fear is a rather silly question. The other day I came across this question on a GCSE maths paper:
x-y = k
x^2+y^2-9=0
These simultaneous equations have exactly one set of solutions.
Show that k = +/_3root2 (I don't know hoe to put the square root sign in.
I have never come across questions like this before and I have no idea what this method/technique is called; I have no idea why it is used and I have no idea about how to solve such questions. I would very much appreciate it if someone could tell me what this technique is called, why it is used and where I might find instructions about how to use it.
Marcus Clayson Answer by math_tutor2020(3816) (Show Source):
Plug this into the other equation to see what happens.
x^2 + y^2 - 9 = 0
x^2 + (y)^2 - 9 = 0
x^2 + (x-k)^2 - 9 = 0
x^2 + x^2 - 2kx + k^2 - 9 = 0
2x^2 - 2kx + k^2 - 9 = 0
In terms of the variable x we have this quadratic template.
ax^2 + bx + c = 0
where in this case,
a = 2
b = -2k
c = k^2 - 9
If the original system has one solution for a fixed specific value of k, then the equation 2x^2 - 2kx + k^2 - 9 = 0 must have one solution.
A quadratic having one solution would only happen when the discriminant is equal to 0.
d = b^2 - 4ac = discriminant
b^2 - 4ac = 0
(-2k)^2 - 4(2)(k^2-9) = 0 .......... plug in the a,b,c values mentioned earlier
4k^2 - 8(k^2-9) = 0
4k^2 - 8k^2 + 72 = 0
-4k^2 + 72 = 0
4k^2 - 72 = 0
4k^2 = 72
k^2 = 72/4
k^2 = 18
k = plus minus sqrt(18)
k = sqrt(18) or k = -sqrt(18)
k = sqrt(9*2) or k = -sqrt(9*2)
k = sqrt(9)*sqrt(2) or k = -sqrt(9)*sqrt(2)
k = 3*sqrt(2) or k = -3*sqrt(2)
Below is a graph of when k = 3*sqrt(2)
This makes the line x-y = k tangent to the circle toward the bottom right portion of the circle.
The circle is centered at (0,0) and has radius 3. GeoGebra and Desmos are two of many graphing tools I recommend to use.
And here is the graph when k = -3*sqrt(2)
We have another situation where the line is tangent to the circle.
That tangent point is the solution to the system.
If -3*sqrt(2) < k < 3*sqrt(2), then the line x-y = k intersects the circle at two different locations.
If k < -3*sqrt(2) or k > 3*sqrt(2), then the line doesn't intersect the circle at all and there are no solutions.
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