SOLUTION: (x+2k)^2 + (y-3k)^2 = 25 pass through the point (1,0) find value of k

Question 1015160: (x+2k)^2 + (y-3k)^2 = 25 pass through the point (1,0)
find value of k
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
(x+2k)^2 + (y-3k)^2 = 25 pass through the point (1,0)
find value of k
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(-2k,3k) is the center of the given circle, call it C.
The distance from (-2k,3k) to (1,0) = 5
--> 25 = (-2k-1)^2 + (3k)^2 = 4k^2 + 4k + 1 + 9k^2
13k^2 + 4k - 24 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1264 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.21356837189471, -1.52126067958701. Here's your graph:

==============
k = the 2 values above.
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There are 2 circles of r = 5 that fit.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
pass through the point (,)
find value of :
....substitute and with coordinates of given point
....solve for

........use quadratic formula

......simplify

exact solutions:

and

approximate solutions:
=>=>=>
and
=>=>

so, you have a circle

and circle

and both should pass through the point (,)
check it:

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