Question

Find the coordinates of the circumcentre of the triangle ABC, whose vertices A, B and C are (4, 6), (0, 4) and (6, 2) respectively.

• A. (2, 4)
• B. (3, 3)
• C. (4, 3)
• D. None of These

Answer 1

The correct option is B (3, 3)

Let the coordinates of circumcentre be O(x, y)

OA = OB = OC = radius

OA = OB


$\Rightarrow {\left(\sqrt{(x-4{)}^2+(y-6{)}^2}\right)}^2={\left(\sqrt{(x-0{)}^2+(y-4{)}^2}\right)}^2$

$\Rightarrow (x-4{)}^2+(y-6{)}^2=(x{)}^2+(y-4{)}^2$

$\Rightarrow x^2+16-8x+y^2+36-12y=x^2+y^2+16-8y$

$\Rightarrow -8x+36-12y=-8y$

$\Rightarrow -8x+36-12y+8y=0$

$\Rightarrow -8x-4y+36=0$

$\Rightarrow -4(2x+y-9)=0$

$\Rightarrow 2x+y-9=0$

$\Rightarrow y=9-2x$ ...(1)

OB = OC = r

$\Rightarrow {\left(\sqrt{(x-0{)}^2+(y-4{)}^2}\right)}^2$

$={\left(\sqrt{(x-6{)}^2+(y-2{)}^2}\right)}^2$

$\Rightarrow (x{)}^2+(y-4{)}^2=(x-6{)}^2+(y-2{)}^2$

$\Rightarrow x^2+y^2+16-8y=x^2+36-12x+y^2+4-4y$

$\Rightarrow 16-8y=40-12x-4y$

$\Rightarrow 16-8y-40+12x+4y=0$

$\Rightarrow -4y+12x-24=0$

$\Rightarrow -4(y-3x+6)=0$

$\Rightarrow y-3x+6=0$

$\therefore y=3x-6$ ... (2)

From (1) & (2), we get

$3x-6=9-2x$

$\Rightarrow 3x+2x=9+6$

$\Rightarrow 5x=15$

$\therefore x=3$

Substituting x = 3 in (1), we get:

$y=9-2x=9-2\times 3=9-6$

$\Rightarrow y=3$

So, the coordinates of the circumcentre is (3, 3).

Hence, option b is correct.