SOLUTION: Show that the circles x^2 + y^2 - 16x - 20y + 115 = 0 and x^2 + y^2 + 8x - 10y + 5 = 0 are tangent and fine the point of tangency

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Question 1087375: Show that the circles x^2 + y^2 - 16x - 20y + 115 = 0 and x^2 + y^2 + 8x - 10y + 5 = 0 are tangent and fine the point of tangency
Found 3 solutions by ikleyn, Fombitz, Boreal:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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The first circle has the center at (x,y) = (8,10) and the radius sqrt%28-115+%2B+64+%2B+100%29 = sqrt%2849%29 = 7 (make completing the squares).


The second circle has the center at (x,y) = (-4,5) and the radius sqrt%28-5+%2B+16+%2B+25%29 = sqrt%2836%29 = 6 (do the same).


The distance between the centers is sqrt%28%288-%28-4%29%29%5E2%2B%2810-5%29%5E2%29 = sqrt%28144+%2B+25%29 = sqrt%28169%29 = 13 = 7 + 6, 

exactly as the sum of the radii.


So, we have external touching in this case.

Solved.


On using completing the squares to transform a general equation of a circle to its standard form see the lessons
    - Standard equation of a circle
    - Find the standard equation of a circle
    - General equation of a circle
    - Transform general equation of a circle to the standard form by completing the squares
    - Identify elements of a circle given by its general equation
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
"Conic sections: Ellipses. Definition, major elements and properties. Solved problems".



Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-16x%2By%5E2+-+20y+%2B+115+=+0
%28x-8%29%5E2%2B%28y-10%29%5E2=+-115%2B64%2B100+
%28x-8%29%5E2%2B%28y-10%29%5E2=49
%28x-8%29%5E2%2B%28y-10%29%5E2=7%5E2
and
x%5E2%2B8x+%2B+y%5E2+-+10y+%2B+5+=+0
%28x%2B4%29%5E2%2B%28y-5%29%5E2=-5%2B16%2B25
%28x%2B4%29%5E2%2B%28y-5%29%5E2=36
%28x%2B4%29%5E2%2B%28y-5%29%5E2=6%5E2
So the point of tangency lies on the line that connects the centers of the circles (8,10) and (-4,5).
If the distance from the centers is x then the point lies %286%2F%286%2B7%29%29 of the distance from (-4,5) to (8,10).
So the x distance from the centers is,
dx=8-%28-4%29=12
So then starting from -4,
x%5Bt%5D=-4%2B12%286%2F13%29=-4%2B72%2F13=-52%2F13%2B72%2F13=20%2F13
And the y distance is,
dy=10-5=5
And starting from 5,
y%5Bt%5D=5%2B5%286%2F13%29=65%2F13%2B30%2F13=95%2F13
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(20%2F13,95%2F13)
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Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + y^2 - 16x - 20y =-115
x^2-16x+64+y^2-20y+100=49, adding 164 to both sides
(x-8)^2+(y-10)^2=7^2
Circle one has a center of (8, 10) and radius 7.
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x^2 + y^2 + 8x - 10y + 5 = 0;
x^2+8x+16+y^2-10y+25=36
(x+4)^2+(y-5)^2=6^2
Circle two has a center of (-4, 5) and radius 6
The tangent line is perpendicular to the line connecting the two radii.
That line has slope 5/12 and its formula is y-y1=m(x-x1); m slope (x1,y1) point. y-10=(5/12)(x-8), or y=(5/12)x+80/12. The slope of the tangent line is -12/5, the negative reciprocal.
The distance between the two centers is 13, so the x component of the tangent point is 6/13 the way from -4 to 8, which has distance 12. That is x=-4+(6/13)*12=-4+72/13, or 20/13.
y=5+(6/13)*5=(65/13)+(30/13)=95/13
the point is at (20/13, 95/13)
the line equation is y-(95/13)=(-12/5)(x-20/13)
This is y=(-12/5)x+(715/65), OR y=(-12/5)x+11

If you set the two circle equations equal to each other, the square terms cancel each other and the result is 24x+10y=110 or 12x+5y=55. The point (20/13, 95/13) is a solution to that equation.