Find the range of value for $k$ for which $kx + y = 3$ meets $x^2 + y^2 = 5$ in two distinct points.
im so stuck can someone give me a clear guide to the correct method and answer, thank you
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1$kx+y=3\implies y=3-kx.$ The line and the circumference have two points in common if $x^2+(3-kx)^2=5$ has two different solutions. – mfl Oct 27 '14 at 13:28
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HINT:
Put $y=3-kx$ in the second equation to form a Quadratic Equation in $x$
Each value of $x$ corresponds to the abscissa of the intersection
Do you know how to find the nature of roots of a Quadratic Equation?
lab bhattacharjee
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Thank you very much :) Once i expand the brackets to $x^2 + 4 - 6kx +kx^2 = 0$ i cant seem to be able to find x still :( – Kelly Oct 27 '14 at 13:30
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@Kelly, You need to arrange as $$Ax^2+Bx+C=0$$ Have you heard of discriminant ? – lab bhattacharjee Oct 27 '14 at 13:37
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Yes i understand the discriminant and everything, i dont understand where i plug $x$ – Kelly Oct 27 '14 at 13:39
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$$x^2+y^2=5$$ $$y=3-kx$$
$$x^2+(3-kx)^2=5$$
$$x^2+9+k^2x^2-6kx=5$$ $$(1+k^2)x^2-6kx+4=0$$
Distinct Real roots $\Rightarrow\Delta\gt0$ $$b^2-4ac\gt0$$ where $b=-6k$ , $a=1+k^2$ , $c=4$
$$36k^2-16(1+k^2)>0$$
$$20k^2-16>0$$ $$\Rightarrow |k|\gt\frac{2}{\sqrt5}$$
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The discriminant is $b^2 - 4ac$ with $b = -6k$, $a = 1 + k^2$, and $c = 4$. – N. F. Taussig Oct 27 '14 at 13:40
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I suggest you to draw a picture, it is easy to see that $-\frac{\sqrt{5}}{2} < k<\frac{\sqrt{5}}{2} $
Paul
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This is a circle with diameter 5 centred at the origin, and a line which is pivoted at y=3. Think of the line as a pendulum sweeping through the plane and intersecting with the circle. Since the pivot is above the top of the circle then during its sweep it will intersect at two distinct points with the circle except when the line becomes tangent to the circle at two points. This is when |k|=2/√5. Note that k is the negative of the slope of the line.
jimalton
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