Question 1202025
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Find the equation of a circle that passes through (2,0) and (8,0) and also touches the Y-axis
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<pre>
The given points A = (2,0) and B = (8,0) are two points in x-axis. They are x-intercept points.


Since the circle passes through these two points, its radius lies on the perpendicular
to x-axis, which bisects the interval between the points A and B.


It means that x-coordinate of the center is x= {{{(2+8)/2}}} = {{{10/2}}} = 5.


Next, since the circle touches y-axis, the radius of the circle is 5 units.


In the circle, we have a cord AB of the length 8-2 = 6; half of the chord has the length 6/2 = 3.
Then y-coordinate of the center is  {{{sqrt(5^2-3^2)}}} = {{{sqrt(25-9)}} = {{{sqrt(16)}}} = 4.


Thus the equation of the circle in standard form is

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

or

    {{{(x-5)^2}}} + {{{(y-4)^2}}} = 25.    <U>ANSWER</U>
</pre>

Solved.