Question

A triangle has vertices at A(2,-2), B(-1, 1) and C(3, 1). Center and radius of the circle in which
the triangle is inscribed
(A) (0, 1), V3 units
(B) (0, – 3), 4 units
(C) (1,1), V6 units
(D) (1,0), 15 units​

Answer 1

a. (0,1)V3units ......

Answer 2

Center of circle: (1,0)

Radius = √5  units

D is correct

Step-by-step explanation:

A triangle has vertices at A(2,-2), B(-1, 1) and C(3, 1)

The triangle ABC is inscribed in a circle. It means all the vertices of triangle is on circle.

The circumcenter of circle would be same as center of circle.

Circumcenter is point of intersection of penpendicular bisector of each side of triangle.

Equation of line Perpendicular bisector of AB passing through mid point of AB.

Point: $(\dfrac{2-1}{2},\dfrac{-1+2}{2})\Rightarrow (\frac{1}{2},-\frac{1}{2})$

Slope of perpendicular bisector line $=-\dfrac{2+1}{-2-1}=1$

Equation of line Perpendicular bisector of AB:

$y+\frac{1}{2}=1(x-\frac{1}{2})$

$y=x-1$  ----------- (1)

Equation of line Perpendicular bisector of BC passing through mid point of BC.

Point: $(\dfrac{-1+3}{2},\dfrac{1+1}{2})\Rightarrow (1,1)$

Slope of perpendicular bisector line $=-\dfrac{3+1}{1-1}=\infty$

Equation of line Perpendicular bisector of BC:

$x=1$  ------------(2)

Center of circle is intersection of eq(1) and eq(2)

y = x - 1  and  x = 1

Center: (1,0)

Radius distance from center to any vertex of triangle.

$\text{Radius }=\sqrt{(2-1)^2+(0+2)^2}=\sqrt{5}$

#Learn more:

Circumcenter of ciricle or perpendicular bisector of line

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