Question

The equation of the circle passing through the point 1, 0 and 0, 1 and having the smallest radius is

Answer 1

Answer:

The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is (a) x2 + y2 + x + y - 2 = 0 (b) x2 + y2 - 2x ...

Answer 2

Step-by-step explanation:

For getting the smallest radius the line joining the points be the diameter of the circle. therefore the midpoint be the center

To find the mid point of   (1,0) and(0,1)

Let (x, y) be the mid point

x = (1 - 0)/2 = 1/2

y = (0 - 1)/2 = -1/2

therefore mid point (x,y) = (1/2, -1/2)

Equation of the square

(x - 1/2)^2 + (y - -1/2)^2 = (x - 1/2)^2 + (y + 1/2)^2

 = x^2 - x + 1/4 + y^2 + y + 1/4

 = x^2 + y^2 -x +y +1/2

Therefore equation of the square is x^2 + y^2

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