Question

If $R=4$ and $r=2$ then the distance between the circum-centre and the in-centre is

• A. $1$
• B. $\sqrt{2}$
• C. $0$
• D. $1.5$

Answer 1

Answer: (C)

$0$

Letting $O$ be the circum-centre centre of triangle $\Delta ABC$ , and $I$ be its in-centre,

and $d$ is distance between in-centre and circum-centre.
Now the extension of $AI$ intersects the circumcircle at $L.$
Then $L$ is the midpoint of arc $BC$.

Join $LO$ and extend it so that it intersects the circumcircle at$M.$

From $I$ construct a perpendicular to$AB$, and let$D$ be its foot,

so $ID=r$.

It is not difficult to prove that triangle $\Delta ADI$ is similar to triangle $\Delta MBL$
so $\frac{ID}{BL}=\frac{AI}{ML}$

i.e.$ID\times ML=AI\times BL$ .......... $ML=2R$
Therefore $2R\times r=AI\times BL$.
Join $BI$.

$\angle BIL=\frac{\angle A}{2}+\frac{\angle ABC}{2}$

$\angle IBL=\frac{\angle ABC}{2}+\angle CBL=\frac{\angle ABC}{2}+\frac{\angle A}{2}$,
$\angle BIL=\angle IBL,$
so $BL=IL$
and $AI\times IL=2R\times r$
Extend $OI$ so that it intersects the circumcircle at P and Q;
then $PI\times QI=AI\times IL=2R\times r$, so$(R+d)\times (R-d)=2R\times r$
i.e.$d^2=R\times (R-2r)$

Now we have$d^2=R\times (R-2r)$

Given $R=4$, $r=2$

Put the value of R and r in above formula

$d=0$

That means in-centre and circum-centre coincide each other.

Hence, the distance between them is $0$.