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

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(52810)   (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 = = 7 (make completing the squares). The second circle has the center at (x,y) = (-4,5) and the radius = = 6 (do the same). The distance between the centers is = = = 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)   (Show Source): You can put this solution on YOUR website!

and

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 of the distance from (-4,5) to (8,10).
So the x distance from the centers is,

So then starting from -4,

And the y distance is,

And starting from 5,

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Answer by Boreal(15235)   (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.

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