Question

The coordinates of the vertices of a triangle are $(0,1),(2,3)$ and $(3,5)$. Find circumcentre and circumradius

• A. $(\frac{9}{2},3),\frac{\sqrt{117}}{2}$
• B. $(14,-6),7\sqrt{5}$
• C. $(-\frac{9}{2},\frac{15}{2}),\frac{5\sqrt{10}}{2}$
• D. $(2,6),\sqrt{29}$

Answer 1

Answer: (C)

$(-\frac{9}{2},\frac{15}{2}),\frac{5\sqrt{10}}{2}$

Let coordinates of circumcentre is $(x,y)$ Then equating distance of circumcentre from vertices we get

$\sqrt{{(x-0)}^2+{(y-1)}^2}=\sqrt{{(x-2)}^2+{(y-3)}^2}=\sqrt{{(x-3)}^2+{(y-5)}^2}$

Solving $\sqrt{{(x-0)}^2+{(y-1)}^2}=\sqrt{{(x-2)}^2+{(y-3)}^2}$

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

$\Rightarrow 4x+4y-12=0\Rightarrow x+y=3$ ...(1)

And solving $\sqrt{{(x-2)}^2+{(y-3)}^2}=\sqrt{{(x-3)}^2+{(y-5)}^2}$

$\Rightarrow x^2-4x+4+y^2-6y+9=x^2-6x+9+y^2-10y+25$

$\Rightarrow 2x+4y-21=0\Rightarrow 2x+4y=21$ ...(2)

From (1) and (2) we get $(x,y)=(-\frac{9}{2},\frac{15}{2})$

And circumradius $r=\sqrt{{(-\frac{9}{2}-0)}^2+{(\frac{15}{2}-1)}^2}=\frac{5\sqrt{10}}{2}$