# The line 3x+4y+k=0 is a tangent to the circle x^2+y^2-8x-10y+32=0. Find 2 possible values of k.Help! I know one should rewrite the 1st equation and substitute it into the equation of the circle...

The line 3x+4y+k=0 is a tangent to the circle x^2+y^2-8x-10y+32=0. Find 2 possible values of k.

Help! I know one should rewrite the 1st equation and substitute it into the equation of the circle (eg. y=-3x-k/4), but I am now very mixed up in the multiplying out part!

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`x^2 + y^2 - 8x + 10y + 32 = 0`

`3x + 4y + k = 0`

The slope of the line 3x+4y+k=0 is -3/4

Writing the circle equation in standard form by completing the square we get

`(x-4)^2 + (y+5)^2 = 9`

The slope of the tangent line of this circle has to be `(-(x-4))/(y+5)`

So where is the slope `= -3/4`

Rewriting the equation as

`(x-4)^2/(y+5)^2 + 1 = 9/(y+5)^2` Substituting `-3/4` for `(-(x-4))/(y+5)` we get

`(-3/4)^2 + 1 = 9/(y+5)^2` we get

`(y+5)^2 = 9/(9/16+1) = 9/(25/16) = 144/25`

Taking the square root we get

`y+5 = +- 12/5`

`y = -5 +- 12/5`

So `y = -37/5` or `y = -13/5`

We can find the x points by using the fact that `(x-4)/(y+5) = 3/4` after I remove the negative sign. For `y=-37/5` we get `x-4 = (3/4)(-12/5) = -9/5` so `x = 11/5`

. For `y=-13/5` we get `x-4=(3/4)(12/5) = 9/5` so `x = 29/5`

At `(11/5, -37/5)` we get `3(11/5)+4(-37/5) + k = 0` gives us `33/5 - 148/5 + k = 0` and finally `k = 115/5 = 23` .

At `(29/5, -13/5)` we get `3(29/5)+4(-13/5) + k = 0` gives us `87/5 - 52/5 + k = 0` and we get `k = -35/5 = -7` . Let's graph to see:

So our answer is k=-7 or k=23

If a line is tangent to a cirlce `(x-a)^2+(y-b)^2 = C` at point (x, y) it has to have slope `-(x-a)/(y-b)` .

Proof:

If the line has equation y=mx+b. We know that the line `y - a = -1/m(x - b)` is perpendicular to y=mx+b and goes through the center (a,b) of the circle. Solving `y - a = -1/m(x-b)` for m and we get `m = -(x-b)/(y-a)` as the slope of the tangent line.