Question 1011293
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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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<pre>
Complete the squares:

{{{x^2 + y^2 - 10x -8y + 25}}} = {{{(x-5)^2}}} + {{{(y-4)^2}}} - {{{16}}}. 

Therefore, the equation 

{{{x^2 + y^2 - 10x -8y + 25}}} = {{{0}}}

is equivalent to the equation

{{{(x-5)^2}}} + {{{(y-4)^2}}} = {{{4^2}}}.

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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<U>Comment from student</U>: but please how did u get the coordinates to be (5,0)?
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<U>My response</U>: 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.
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