SOLUTION: determine the equation of the circle whose center is at (4,5) and tangent to the circle whose equation is x^2+y^2+4x+6y-23=0.

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Question 534153: determine the equation of the circle whose center is at (4,5) and tangent to the circle whose equation is x^2+y^2+4x+6y-23=0.
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
determine the equation of the circle whose center is at (4,5) and tangent to the circle whose equation is x^2+y^2+4x+6y-23=0
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Find the center of x^2+y^2+4x+6y-23=0
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x^2+y^2+4x+6y = 23


Center at (-2,-3), r = 6
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Find the distance from the center to (4,5)

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The radius of the 2nd circle is 4, 10-6
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Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
The given circle equation can be written as

and completing squares as
or
showing that its radius is 6 and its center is (-2, -3).
The distance between the centers is

If the circles meet in between the centers (they are externally tangent), the radius of the second circle will be

and the equation for the second circle will be

which can be written as

If we make the second circle contain the first one (internally tangent circles), then the radius would be

and the equation for the second circle would be

which can be written as
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