SOLUTION: Please help me with this! Show that the two circles (x - 3)^2 + (y-4)^2 = 25 and (x-1)^2 + (y-5/2)^2 =225/4. touch each other.
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Question 992326: Please help me with this! Show that the two circles (x - 3)^2 + (y-4)^2 = 25 and (x-1)^2 + (y-5/2)^2 =225/4. touch each other. Found 3 solutions by Edwin McCravy, ikleyn, Theo:Answer by Edwin McCravy(20054) (Show Source):
Subtract the two equations to get the equation of the line through
their point(s) of intersection if they have any points in common.
Factor first two and last two terms on the left as the
difference of squares:
Multiply through by 4
Divide through by -4
This is the equation of the line through their
point(s) of intersection if the intersect.
Substitute in
Multiply through by 9
Divide through by 25
That give the double solution x=7. The double solution
means that the circles intersect at one point, where x=7.
Substituting in
Therefore they intersect at the one point (7,7)
Edwin
You can put this solution on YOUR website! .
Please help me with this! Show that the two circles (x - 3)^2 + (y-4)^2 = 25 and (x-1)^2 + (y-5/2)^2 =225/4. touch each other.
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The first equation is for the circle of the radius r = 5 with the center at the point (x,y) = (3,4).
The second equation is for the circle of the radius R = = with the center at the point (x,y) = (1, ) = (1, 2.5).
The distance between the centers is d = = = = = .
Notice that the second circle is larger than the first one, and the center of the first circle is located
inside of the second circle. See the Figure below.
Now notice that R = d + r: 7.5 = 2.5 + 5.
It means that the first circle touches the second one from the inside.