Question 803155: Find the equation of circle which touches x-axis at (3,0) and passes through the point (1,2).
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website! A strategy could be, find the line that the two given points define. The line contains the center of the desired circle. The equation for this line is (using formula for slope, and then starting from point-slope formula), y=-x+3.
This means that the center point for your desired circle is at some general (x, -x+3).
USE THE DISTANCE FORMULA between (x, -x+3) and each of the two given points; and equate the distances and solve for x. You may find two values. One will make sense and one will not make sense. WHY? Because you want the center which would agree with CONTAINING (1,2) and only TOUCHES (3,0). The circle center should be in quadrant 1.
Distance from center to (3,0) will then be the radius. Fill in the standard form,  , center being (h,k).
I have not worked all the way through this, but only described a strategy to solve the problem.
Note: I suspect the center will also be contained on the vertical line, x=3. I say this because of "... touches the x axis..."
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