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# Two identical circles touch at the point P(9,3)

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Two identical circles touch at the point P(9,3)

one of the circles has equation x^2 + y^2 - 10x - 4y +12 = 0

find the equation of the other circle

Oct 27, 2018

#1
+3

Lets figure out the standard equation for a circle by completing the square

x^2-10x+25    + y^2-4x+4  = -12 +25 + 4

9x-5)^2  + (y-2)^2 = 17                                     center is h, k  = 5,2

We want a straight line frome center through 9,3  to new center (so the circles will be tangent at 9,3)

5,2   to  9, 3    is  a change of  4,1      add that to 9, 3   to get    13,4

SO the identical tangent circle is

(x-13)^2 + (y-4)^2 = 17

Oct 27, 2018

#1
+3

Lets figure out the standard equation for a circle by completing the square

x^2-10x+25    + y^2-4x+4  = -12 +25 + 4

9x-5)^2  + (y-2)^2 = 17                                     center is h, k  = 5,2

We want a straight line frome center through 9,3  to new center (so the circles will be tangent at 9,3)

5,2   to  9, 3    is  a change of  4,1      add that to 9, 3   to get    13,4

SO the identical tangent circle is

(x-13)^2 + (y-4)^2 = 17

ElectricPavlov Oct 27, 2018
#2
+3

Here is a graph: Oct 27, 2018