Question

the equation of the circle passing through 2, 0 and 0, 4 and having the minimum radius is​

Answer 1

Answer:

The circle passes through A ( 2 , 0 ) and B ( O ,

4 ) . There will be infinite number of circles passing through these points . DrawAB . 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 ABis the diameter .

So AB = / ( 2 + 4 ) = 2 / 5 So radius = 75 .

Area of the smallest circle = 5 11 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Long way :

( x - a ) 2 + ( y - b ) = 12 ( 2 - a ) + b * = r2 a + ( 4 - b ) = r = > a2 - 4 a + 4 + b * = r2 = > a + b2 - 8 b + 16 = r2 Solving these equations we get a = 2b - 3 .

Substitute it in the first equation ,

b2 - 4b + 5 = 1 / 5 = P ( b ) = y dy / db = 2 b - 4 = 0 = > b = 2 .

a minimum value when b = 2 . P ( b ) = y has

Minimum value = P ( 2 ) = 4 - 8 + 5 = 1 So smallest radius = r = -

15 Area of smallest circle = 511

Answer 2

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