Question

Find the equation of circumcircle of a triangle whose vertices are (1,0) , (-1,0) and (0,1).

Answer 1

Answer: $x^{2} +y^{2} -y-1=0.$

Step-by-step explanation: Let, the equation of the circumcircle of the triangle with vertices A(1,0), B(-1,0) and C(0,1) be

$(x-a)^{2} +(y-b)^{2}=r^{2}$, where (x,y) is a point on the circle and r is the radius.

Surely, the three vertices of the triangle will satisfy the equation of the circle.

So,

$(1-a)^{2} +(0-b)^{2}=r^{2},~(-1-a)^{2} +(0-b)^{2}=r^{2}\\~\textup{and}~(0-a)^{2} +(1-b)^{2}=r^{2}.$

Comparing these equations with one another, we have

$a=0, b=\dfrac{1}{2} ~\textup{and}~ r=\dfrac{\sqrt{5} }{2}.$

Thus, putting these values in the equation of the circle, we find

$x^{2} +y^{2} -y-1=0.$