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
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
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?
$$x^2+y^2=5$$ $$y=3-kx$$
$$x^2+(3-kx)^2=5$$
$$x^2+9+k^2x^2-6kx=5$$ $$(1+k^2)x^2-6kx+4=0$$
Distinct Real roots $\Rightarrow\Delta\gt0$ $$b^2-4ac\gt0$$ where $b=-6k$ , $a=1+k^2$ , $c=4$
$$36k^2-16(1+k^2)>0$$
$$20k^2-16>0$$ $$\Rightarrow |k|\gt\frac{2}{\sqrt5}$$
I suggest you to draw a picture, it is easy to see that $-\frac{\sqrt{5}}{2} < k<\frac{\sqrt{5}}{2} $
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.