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?

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Feb 9, 2018

Answer:

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 #3x-2y+k=0#, which gives us #y=3/2x+k/2# and putting this in the equation of circle #x^2+y^2-4x-9=0#, we get

#x^2+((3x)/2+k/2)^2-4x-9=0#

or #x^2+(9x^2)/4+k^2/4+3/2kx-4x-9=0#

or #4x^2+9x^2+k^2+6kx-16x-36=0#

or #13x^2+(6k-16)x+k^2-36=0#

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

#(6k-16)^2-52(k^2-36)>0#

or #36k^2-192k+256-52k^2+1872=0#

or #-16k^2-192k+2128>0#

or #k^2+12k-133<0#

or #(k+19)(k-7)<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#

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