question_answer

Let A, B, C be the angles of \[\Delta ABC\] with vertex \[A\left( 4,-1 \right)\] and \[x-1=0\] and \[x-y=1\]are internal angle bisectors through B and C respectively. Let D, E, F be points of contact of sides BC. CA and AB with incircle of \[\Delta ABC\]. If D?, E?, F' are images of D, E and F in internal angle bisector of A, B, C, then equation of circumcircle of \[\Delta D'\,E'\,F\] is

A) \[{{\left( x-1 \right)}^{2}}+{{y}^{2}}=5\]

B) \[{{x}^{2}}+{{\left( y-1 \right)}^{2}}=25\]

C) \[{{(x-1)}^{2}}+{{\left( y-1 \right)}^{2}}=5\]

D) \[{{x}^{2}}+{{y}^{2}}=25\]

Correct Answer: A

Solution :

Mirror image of A \[(4,-1)\] with espect to line \[x-1=0\] and \[x-y-1=0\] ire respectively \[(-2,-1)\] and (0, 3) which lie on BC

Equation of \[BC=2x-y+3=0\]

\[r=\left| \frac{2-0+3}{\sqrt{5}} \right|=\frac{5}{\sqrt{5}}=\sqrt{5}\]

Now image of D, E , F are also lies on circumcircle with respect to the diameter.

\[\therefore \]    D', E', F' are lie on in circle of \[\Delta ABC\]

Equation of in circle of

\[\Delta ABC={{\left( x-1 \right)}^{2}}+{{y}^{2}}=5\]