Question

Let A, B, C be angles of triangle ABC with vertex A $(4-1)$ and $x-1=0$ and $x-y-1=0$ are internal angle bisectors through B and C respectively. Let D, E, F be points of contact of sides bisector of A, B, C, then equation of circumcircle of triangle DEF is

• A. ${(x-1)}^2+y^2=5$
• B. $x^2+{(y-1)}^2=25$
• C. ${(x-1)}^2+{(y-1)}^2=5$
• D. $x^2+y^2=25$

Answer 1

The correct option is A ${(x-1)}^2+y^2=5$

mirror images of A(4,-1) about the two bisectors $x-y-1=0$ and $x-1=0$ are respectively (0, 3) and (-2, -1) which lie in BC
$\therefore$ Equation of BC is $2x-y+3=0\Rightarrow$ inradius $r=\frac{5}{\sqrt{5}}=\sqrt{5}$
Again image of a point circumcircle with respect to diameter lies on circle again i.e circle is $(x-1{)}^2+(y-0{)}^2=(\sqrt{5}{)}^2$