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

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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) About Me  (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 ax%5E2%2Bbx%2Bc=0 (in our case 13x%5E2%2B4x%2B-24+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284%29%5E2-4%2A13%2A-24=1264.

Discriminant d=1264 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+1264+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+1264+%29%29%2F2%5C13+=+1.21356837189471
x%5B2%5D+=+%28-%284%29-sqrt%28+1264+%29%29%2F2%5C13+=+-1.52126067958701

Quadratic expression 13x%5E2%2B4x%2B-24 can be factored:
13x%5E2%2B4x%2B-24+=+%28x-1.21356837189471%29%2A%28x--1.52126067958701%29
Again, the answer is: 1.21356837189471, -1.52126067958701. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+13%2Ax%5E2%2B4%2Ax%2B-24+%29

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k = the 2 values above.
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There are 2 circles of r = 5 that fit.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B2k%29%5E2+%2B+%28y-3k%29%5E2+=+25 pass through the point (1,0)
find value of k:
%28x%2B2k%29%5E2+%2B+%28y-3k%29%5E2+=+25....substitute x and y with coordinates of given point
%281%2B2k%29%5E2+%2B+%280-3k%29%5E2+=+25....solve for k
1%2B4k%2B4k%5E2+%2B+9k%5E2+=+25
13k%5E2%2B4k+=+25-1
13k%5E2%2B4k+-24=0........use quadratic formula
k+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
k+=+%28-4+%2B-+sqrt%28+4%5E2-4%2A13%2A%28-24%29+%29%29%2F%282%2A13%29+
k+=+%28-4+%2B-+sqrt%28+16%2B1248+%29%29%2F26+
k+=+%28-4+%2B-+sqrt%28+1264+%29%29%2F26+
k+=+%28-4+%2B-+sqrt%28+16%2A79+%29%29%2F26+
k+=+%28-4+%2B-+4sqrt%28+79+%29%29%2F26+......simplify
k+=+%28-2+%2B-+2sqrt%28+79+%29%29%2F13+
k+=+%28-2%2F13+%2B-+2sqrt%28+79+%29%2F13+%29
exact solutions:
k+=+-2%2F13+%2B+2sqrt%28+79+%29%2F13+
and
k+=+-2%2F13+-+2sqrt%28+79+%29%2F13+
approximate solutions:
k+=+-2%2F13+%2B+2sqrt%28+79+%29%2F13+=>k+=+-0.1538461538461538+%2B+0.1538461538461538%288.9%29+=>k=1.21538461538461502=>k=1.22
and
k+=+-2%2F13+-+2sqrt%28+79+%29%2F13+=>k+=+-0.1538461538461538+-0.1538461538461538%288.9%29++=>k=+-1.523

so, you have a circle
%28x%2B2%2A1.22%29%5E2+%2B+%28y-3%2A1.22%29%5E2+=+25
%28x%2B2.44%29%5E2+%2B+%28y-3.66%29%5E2+=+25
and circle
%28x%2B2%2A%28-1.523%29%29%5E2+%2B+%28y-3%2A%28-1.523%29%29%5E2+=+25
%28x-3.046%29%5E2+%2B+%28y%2B4.569%29%5E2+=+25
and both should pass through the point (1,0)
check it: