SOLUTION: the equation of a circle is given by x^2 + y^2 - 10x -8y + 25 = o i. show that the circle touches the x-axis ii. find the coordinates of the point of contact

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: the equation of a circle is given by x^2 + y^2 - 10x -8y + 25 = o i. show that the circle touches the x-axis ii. find the coordinates of the point of contact      Log On


   

Question 1011293: the equation of a circle is given by x^2 + y^2 - 10x -8y + 25 = o
i. show that the circle touches the x-axis
ii. find the coordinates of the point of contact
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
The x-axis is the line y=0. You expect exactly ONE value for x when y=0.

x%5E2+%2B+y%5E2+-+10x+-8y+%2B+25+=+0
x%5E2%2B0-10x-0%2B25=0
x%5E2-10x%2B25=0
%28x-5%29%28x-5%29=0---------the same binomial factor, squared.

highlight%28x=5%29------the only solution for y=0.

Answer by ikleyn(52799) About Me  (Show Source):
You can put this solution on YOUR website!
.
the equation of a circle is given by x^2 + y^2 - 10x -8y + 25 = o
i. show that the circle touches the x-axis
ii. find the coordinates of the point of contact
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Complete the squares:

x%5E2+%2B+y%5E2+-+10x+-8y+%2B+25 = %28x-5%29%5E2 + %28y-4%29%5E2 - 16. 

Therefore, the equation 

x%5E2+%2B+y%5E2+-+10x+-8y+%2B+25 = 0

is equivalent to the equation

%28x-5%29%5E2 + %28y-4%29%5E2 = 4%5E2.

This is the equation of the circle with the center at the point (x,y) = (5,4) and the radius of 4 units.

Since y-coordinate of the center is 4 and the radius of the circle equals 4 too, the circle touches the x-axis.

The coordinate of the contact point is (5,0).


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Comment from student: but please how did u get the coordinates to be (5,0)?
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My response: The center of the circle is at the point (5,4) and the radius is 4.
Obviously, the circle touches x-axis and the contact point is (5,0).
Zero is the y-coordinate of the contact point.
Make a sketch.