Question

For a triangle $ABC$ with vertices as $A(5,-1),B(-1,5),C(6,6)$, the coordinates of circumcentre and value of circumradius (in units) is

• A. $(\frac{23}{8},\frac{23}{8}),\frac{25\sqrt{2}}{8}$
• B. $(\frac{23}{8},\frac{8}{23}),\frac{25}{184}$
• C. $(0,0),\sqrt{26}$
• D. $(1,1),\sqrt{20}$

Answer 1

The correct option is A $(\frac{23}{8},\frac{23}{8}),\frac{25\sqrt{2}}{8}$


Let $O(\alpha ,\beta )$ be the circumcentre of $△ABC$
$D\equiv (\frac{6-1}{2},\frac{6+5}{2})\equiv (\frac{5}{2},\frac{11}{2})$
$E\equiv (\frac{6+5}{2},\frac{6-1}{2})\equiv (\frac{11}{2},\frac{5}{2})$

$F\equiv (\frac{5-1}{2},\frac{-1+5}{2})\equiv (2,2)$
$\because OD\perp BC$
$\Rightarrow m_{OD}\times m_{BC}=-1$
$\frac{\frac{11}{2}-\beta }{\frac{5}{2}-\alpha }\times \frac{6-5}{6-(-1)}=-1$

$\Rightarrow 11-2\beta =-35+14a$

$\Rightarrow 7\alpha +\beta =23\cdots \left(i\right)$
$\because OE\perp AC$

$\Rightarrow m_{OE}\times m_{AC}=-1$

$\frac{\frac{5}{2}-\beta }{\frac{11}{2}-a}\times \frac{6-(-1)}{6-5}=-1$

$\Rightarrow 7\beta +\alpha =23\cdots \left(ii\right)$
By solving $\left(i\right)$ and $\left(ii\right)$
We get circumcentre $=(\frac{23}{8},\frac{23}{8})$

Circumradius $=OC=OB=OA$

$=\sqrt{{(\frac{23}{8}-6)}^2+{(\frac{23}{8}-6)}^2}=\frac{25\sqrt{2}}{8}$ units