Question

The equation of the circle passing through (2, 0) and (0, 4) and having minimum radius has an area
(a) (b)   (c) (d) 

Answer 1

This can be done in two ways.  Quick and simple way is:

The circle passes through A(2,0) and B(0,4).  There will be infinite number of circles passing through these points.  Draw AB. If center lies outside AB, then AB is chord on the circle. Then the diameter of circle is bigger than AB.

If the center of circle is the midpoint of AB and AB is the diameter.  So 
     AB = √(2²+4²) = 2√5           So radius = √5.

      Area of the smallest circle = 5 π
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Long way:
   (x-a)² + (y-b)² = r²
    (2 - a)² + b² = r²        =>  a² - 4 a + 4 + b² = r²
  a² + (4-b)² = r²        =>    a² + b² - 8 b +16 = r²   

Solving these equations we get a = 2b - 3.     Substitute it in the first equation,
    
         b² - 4b + 5 = r²/5   = P(b) = y 
 
   d y / d b = 2 b - 4   = 0   =>  b = 2 .   P(b) = y has a minimum value when b = 2.
 
         Minimum value = P(2) = 4 - 8 + 5 = 1

So smallest radius = r = √5

Area of smallest circle  = 5π