Question

The line `3x+4y-k=0` is tangent to the circle `x^2+y^2=16`. What are the possible values of k?

Answer 1

`k=+-20`

Explanation:

Let us find points of intersection of line `3x+4y-k=0` and circle `x^2+y^2=16`. We can do this by putting value of `y` from first equation i.e. `y=(k-3x)/4` and we get

`x^2+(k-3x)^2/16=16`

or `16x^2+k^2+9x^2-6kx=256`

i.e. `25x^2-6kx+k^2-256=0`

This would give two values of `x` and corresponding two values of `y` i.e. two points. But tangent cuts the circle in only at one point. This will be so when discriminant is zero i.e.

`(-6k)^2-4*25*(k^2-256)=0`

or `-64k^2+25600=0` or `k=+-20`

graph{(x^2+y^2-16)(3x+4y-20)(3x+4y+20)=0 [-10, 10, -5, 5]}

Answer 2

`k=+-20`.

Explanation:

We know from Geometry, that, the `bot"-distance"` from the

Centre of a Circle to its tangent line equals the

Radius of the Circle.

Let us use this fact to solve the Problem :

We see that, for the circle `S : x^2+y^2=16`, its centre is

`O=O(0,0)`, and radius `r=4`.

`"But, the" bot"-dist. p from "O" to the tgt. : "3x+4y-k=0`, is,

`p=|3(0)+4(0)-k|/sqrt(3^2+4^2)=|k|/5`.

Since, `p=r rArr |k|/5=4 rArr k=+-20`.