Question

Two of the vertices of an equilateral triangle are $(0,4)$ and $(0,6)$ and the third vertex lies in the second quadrant. Then the equation of the incircle of the triangle is

• A. ${(x-\frac{5}{2})}^2+(y-5{)}^2=\frac{29}{4}$
• B. ${(x+\frac{5}{2})}^2+(y-5{)}^2=\frac{1}{3}$
• C. $x^2+y^2+\frac{2}{\sqrt{3}}x-10y+25=0$
• D. ${(x+\frac{\sqrt{3}}{2})}^2+(y-5{)}^2=\frac{1}{3}$

Answer 1

The correct option is C $x^2+y^2+\frac{2}{\sqrt{3}}x-10y+25=0$

Let A is other vertex of equiateral triangle ABC
$\therefore AliesonmedianofBc$
so, A(x,s)
$In△ACM,$
$\therefore Am=Accos{30}^0$
$Am=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$
$\therefore A=(-\sqrt{3},5)$
for equilateral friangle incentre and cenhoid are coincide of same point
$so,incenterof△ABCi{s}_o(\frac{-\sqrt{3}}{3},\frac{15}{3})$
$0\equiv (-\frac{1}{\sqrt{3}},5)$
so, equation of in circle is
$(x+\frac{1}{\sqrt{3}}{)}^2+(y-5{)}^2=(oA{)}^2=(\frac{1}{\sqrt{3}}{)}^2$
$\Rightarrow x^2+y^2+\frac{2x}{\sqrt{3}}-10y+25=0$