Question

The number of incircles that touch all the straight lines
$x$ $+$ $y$ $=$ $2$, $x$ $-$ $y$ $+$ $4$ $=$ $0$ and $y=3$ will be

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

$0$

If a circle,with center(let$(h,k)$) touches all three given lines,its distances i.e.,distance of center of circle from each of the lines must be same.

$L1:x+y=2$

$L2:x-y=-4$

$L3:y=3$

$\therefore distanceof(h,k)fromL1;$

$\frac{|h+k-2|}{\sqrt{2}}=k1$

$distanceof(h,k)fromL2;$

$\frac{|h-k+4|}{\sqrt{2}}=k2$

$distanceof(h,k)fromL3;$

$|k-3|=k3$

$k1=k2=k3,ifincircletouchesallthelines.$

$k1=k2$

$\frac{h+k-2}{\sqrt{2}}=\frac{h-k+4}{\sqrt{2}}$

$⟹2k=6$

$⟹k=3$

$ifk=3$

$k3=|k-3|=0$

$\therefore Thiscaseisnotpossible.$

$Case2:$

$\frac{h+k-2}{\sqrt{2}}=\frac{-(h-k+4)}{\sqrt{2}}$

$⟹h+k-2=-h+k-4$

$⟹2h=-2$

$⟹h=-1$

$⟹\frac{-1+k-2}{\sqrt{2}}=|k-3|$

$implies-1+k-2=k\sqrt{2}-3\sqrt{2}$

$⟹k(\sqrt{2}-1)=3(\sqrt{2}-1)$

$⟹k=3$

$againifk=3,$

$k3=|k-3|=0,$

$\therefore Thiscaseisnotpossible.$

$Case3:$

$\frac{-1+k-2}{\sqrt{2}}=-(k-3)$

$implies-3+k=-k\sqrt{2}+3\sqrt{2}$

$⟹k(\sqrt{2}+1)=3(\sqrt{2}+1)$

$⟹k=3$

$again,k3=0$

$D)0$