Question

Find the equation and centre of circle passing through the points (3,4),(3,2),(1,4)

Answer 1

Centre : (2,3)
Radius : √2
Equation :
${x}^{2} + {y}^{2} - 4x - 6y + 11 = 0$

Answer 2

$(x-2)^2+(y-3)^2=2$ is the required equation

Step-by-step explanation:

The center of the circle is located on the perpendicular bisector of each side of the triangle with vertices (3,4), (3,2), and (1,4).

• y=3 is the perpendicular bisector of their line segment ((3,4) to (3,2))

• x=2 is the perpendicular bisector of the line segment ((3,4) to (1,4)).

Centre = (2,3)

• r= distance from (2,3) to any of the 3 given points - Using (3,4)

$r^2=(3-2)^2+(4-3)^2=2$

Equation of circle: $(x-2)^2+(y-3)^2=2$