SOLUTION: find the equation of the circle concentric with the circle x^2+y^2-4x+6y-17=0 which has a tangent of 3x-4y+7=0.

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Question 1191482: find the equation of the circle concentric with the circle x^2+y^2-4x+6y-17=0 which has a tangent of 3x-4y+7=0.
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the equation of the concentric circle to the circle x^2+y^2-4x+6y-17=0 which has a tangent of 3x-4y+7=0.
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By completing the square method, find the center of the circle.  It is the point (2,-3).


To find the radius of the required circle, find the distance from the center (2,-3) to the given line 3x-4y+7=0.


For finding the distance from a given point to a straight line, there is a remarkable formula.


    Let the straight line in a coordinate plane is defined in terms of its linear equation 

         a*x + b*y + c = 0,

    where "a", "b" and "c" are real numbers, and let P = (x%5B0%5D,y%5B0%5D) be the point in the coordinate plane. 

    Then the distance from the point P to the straight line is equal to

        d = abs%28a%2Ax%5B0%5D+%2B+b%2Ay%5B0%5D+%2B+c%29%2Fsqrt%28a%5E2+%2B+b%5E2%29.


Regarding this formula, see the lesson
    The distance from a point to a straight line in a coordinate plane
in this site.


Your straight line is 3x - 4y + 7 = 0.


Substitute the given data  a= 3, b= -4, c= 7,  x%5B0%5D = 2,  y%5B0%5D= -3  into the formula to get the distance under the question


    abs%283%2A2+%2B+%28-4%29%2A%28-3%29+%2B+7%29%2Fsqrt%283%5E2%2B%28-4%29%5E2%29 = abs%2825%29%2Fsqrt%2825%29 = 25%2F5 = 5.


Thus the radius of the circle should be 5 units.


Then the standard form equation of the circle is


    %28x-2%29%5E2 + %28y%2B3%29%5E2 = 25.      ANSWER

Solved.



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
find the equation of the circle concentric with the circle x^2+y^2-4x+6y-17=0 which has a tangent of 3x-4y+7=0.
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Find the center of the circles:
x^2+y^2-4x+6y-17=0
x^2+y^2-4x+6y = 17
(x^2 - 4x + 4) + (y^2 + 6y + 9 = 17 + 4 + 9 = 30
(x-2)^2 + (y+3)^2 = 30
Center at (2,-3)
====================
Find the distance from the center to the line:
The slope of the line is 3/4
The slope of lines perpendicular is -4/3
----
The line thru (2,-3) with a slope of -4/3 ---> y+3 = (-4/3)*(x-2)
y + 3 = -4x/3 + 8/3
3y = -4x - 1
4x + 3y = -1
----
Find the intersection with 3x-4y+7=0
3x-4y = -7
4x+3y = -1
-------
9x - 12y = -21
16x +12y = -4
---------------------- Add
25x = -25
x = -1
-3 - 4y = -7
-4y = -4
y = 1
---> the intersection is (-1,1)
Find the distance from the center at (2,-3) to (-1,1)
d^2 = diffy^2 + diffx^2 = 4^2 + 3 = 25
---> the circle is (x-2)^2 + (y+3)^2 = 25