Question

The coordinates of the vertices of a triangle are (1,2),(2,3),(3,1). Find the coordinates of the centre of its circumcircle and the circumradius.

• A. Circumcentre $(\frac{13}{6},\frac{11}{6})$
• B. Circumcentre $(\frac{11}{6},\frac{13}{6})$
• C. Circumradius = $\frac{5\sqrt{2}}{6}$
• D. Circumradius = $\frac{3\sqrt{5}}{6}$

Answer 1

Answer: (A), (C)

The correct options are
A

Circumcentre $(\frac{13}{6},\frac{11}{6})$

C

Circumradius = $\frac{5\sqrt{2}}{6}$

Let A(1,2), B(2,3), C(3,1) be the vertices of $\Delta ABC$ and P(x,y) be the circumcentre of $\Delta ABC.$

PA =PB =PC (radii of same circle)

$P{A}^2=P{B}^2$

$(x-1{)}^2+(y-2{)}^2=(x-2{)}^2+(y-3{)}^2$

$\Rightarrow x^2-2x+1+y^2-4y+4=x^2-4x+4+y^2-6y+9$

$\Rightarrow -2x-4y+5=-4x-6y+13$

$\Rightarrow 2x+2y=8$

$\Rightarrow x+y=4,$

x =4 -y .......(i)

Again,

$P{B}^2=P{C}^2$

$(x-2{)}^2+(y-3{)}^2=(x-3{)}^2+(y-1{)}^2$

$x^2-4x+4+y^2-6y+9=x^2-6x+9+y^2-2y+1$

$\Rightarrow -4x-6y+13=-6x-2y+10$

$\Rightarrow 2x-4y=-3$

$\Rightarrow 2(4-y)-4y=-3$, from (i)

$\Rightarrow 8-2y-4y=-3\Rightarrow 6y=11$

$\therefore y=\frac{11}{6}$

From (i), $x=4-\frac{11}{6}=\frac{13}{6}$

Circumcentre $(\frac{13}{6},\frac{11}{6})$

Circumradius = $PA=\sqrt{{(\frac{13}{6}-1)}^2+{(\frac{11}{6}-2)}^2}$

$=\sqrt{{\left(\frac{7}{6}\right)}^2+{(-\frac{1}{6})}^2}$

$=\sqrt{\left(\frac{50}{36}\right)}=\frac{5\sqrt{2}}{6}$.