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 BC, CA and AB with incircle of $Δ A B C$ . If $D^{'}, E^{'}, F^{'},$ are images of D, E and F through internal angle bisectors of A, B, C respectively then which of the following options(s) is/are TRUE?

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Equation of incircle of $Δ A B C$ is $x^{2} + y^{2} = 25$

Equation of incircle of $Δ A B C$ is ${(x - 1)}^{2} + y^{2} = 5$ $Δ D' E' F'$

Area of $Δ A B C$ is 23 sq. units

Circumcircle of is ${(x - 1)}^{2} + y^{2} = 5$

answer is A, B.

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Detailed Solution

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
$∴$ Equation of BC is $2 x - y + 3 = 0 \Rightarrow$ In radius $r = \frac{5}{\sqrt{5}} = \sqrt{5}$
Again image of a point on circumcircle with respect to diameter lies on circle again i.e. circle is ${(x - 1)}^{2} + {(y - 0)}^{2} = {(\sqrt{5})}^{2}$