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Question 1180823: When k is a constant, find k for which the length of perpendicular dropped from point (2, 1) to line kx + y + 1 = 0 is √3.
Note: Can you please show your full solution? Thank you!
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52809)   (Show Source):
You can put this solution on YOUR website! .
When k is a constant, find k for which the length of perpendicular dropped from point (2, 1) to line kx + y + 1 = 0 is √3.
Note: Can you please show your full solution? Thank you!
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To make the solution easier, I re-formulate the problem in THIS EQUIVALENT WAY:
Find the coefficient "k" such that the line kx + y + 1 = 0
is tangent to the circle of the radius  centered at the point (2,1)
Solution
The standard form equation of the circle is
 +  = 3. (1)
The equation of the line is
y = -kx - 1. (2)
Substitute the equation (2) into equation (1), replacing y in the equation (1).
Doing this way, you will get then a single equation for only one unknown x
+  = 3 (3)
or
+  = 3. (4)
The idea of the solution to the problem is to reduce equation (4) to standard form quadratic equation
and then to find the value of "k" from the condition that this equation has ONLY ONE real solution
(which is equivalent to the fact that the line(2) and the circle (1) have ONLY ONE common point).
So, we simplify equation (4) step by step
x^2 - 4x + 4 + k^2x^2 + 4kx + 4 = 3
x^2(1+k^2) - 4x(1-k) + 5 = 0. (5)
Now we will find "k" from the condition that the discriminant of equation (5) is equal to zero.
The discriminant d is
d = b^2 - 4ac =  -  =  -  .
The condition d = 0 is
=
4(1-k)^2 = 5(1+k^2)
4 - 8k + 4k^2 = 5 + 5k^2
k^2 + 8k + 1 = 0
 =  =  =  .
So, there are 2 (two, TWO) values for k:  =  and  =  .
ANSWER. There are two solutions for "k" : = and = .
Solved, answered and carefully explained in all details.
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Answer by greenestamps(13200)  (Show Source):
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