# The line 3x-2y+k=0, where k is a constant, intersects the curve x^2 + y^2 -4x-9=0 at two points. Find the range of k?

Feb 9, 2018

Range of $k$ is $- 19 < k < 7$

#### Explanation:

To find points of intersection of a line and curve, we should solve them as simultaneous equation.

Here we have a linear equation indicating a line and a quadratc equation indicating a conic section (actually a circle). When we substitute one variable in terms of another variable from linear equation in to quadratic equation, we get a quadratic equation in other variable.

If the quadratic equation has two roots, they intersect at two points and if it has one root, it is a tangent. In case it has no roots, it means line does not intersect the conic section at all.

Here we have $3 x - 2 y + k = 0$, which gives us $y = \frac{3}{2} x + \frac{k}{2}$ and putting this in the equation of circle ${x}^{2} + {y}^{2} - 4 x - 9 = 0$, we get

${x}^{2} + {\left(\frac{3 x}{2} + \frac{k}{2}\right)}^{2} - 4 x - 9 = 0$

or ${x}^{2} + \frac{9 {x}^{2}}{4} + {k}^{2} / 4 + \frac{3}{2} k x - 4 x - 9 = 0$

or $4 {x}^{2} + 9 {x}^{2} + {k}^{2} + 6 k x - 16 x - 36 = 0$

or $13 {x}^{2} + \left(6 k - 16\right) x + {k}^{2} - 36 = 0$

It will have two roots if discriminant is greater than zero i.e.

${\left(6 k - 16\right)}^{2} - 52 \left({k}^{2} - 36\right) > 0$

or $36 {k}^{2} - 192 k + 256 - 52 {k}^{2} + 1872 = 0$

or $- 16 {k}^{2} - 192 k + 2128 > 0$

or ${k}^{2} + 12 k - 133 < 0$

or $\left(k + 19\right) \left(k - 7\right) < 0$

This is so, when either $k + 19 > 0$ and $k - 7 < 0$ i.e. $- 19 < k < 7$

or $k + 19 < 0$ and $k - 7 > 0$ i.e. $k < - 19$ and $k > 7$, but this is not possible.

Hence solution is $- 19 < k < 7$