Question

Equation of the circle passing through (1,0) and(0,1) and having the smallest possible radius is

Answer 1

Answer:

equation of the square is x^2 + y^2 -x +y +1/2

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 -x +y +1/2

Answer 2

Answer:

Step-by-step explanation:

 Required circle passes through A(1,0)& B(0,1).

For a circle having the smallest possible radius AB should be the diameter of the circle.

Let P(x,y) be any point on the circumfrence of the circle. Angle APB =90°

Slope of AP(m1) =(y-0)/(x-1)=y/(x-1)

Slope of BP(m2)=(y-1)/(x-0)=(y-1)/x

m1×m2 = -1

y/(x-1)×(y-1)/x= -1

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

y^2-y=-x^2+x

x^2+y^2-x-y=0 , Answer