Question

A right angled isosceles triangle is inscribed in the circle $x^2+y^2-4x-2y-4=0$ then length of the side of the triangle is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $3\sqrt{2}$
• D. $4\sqrt{2}$

Answer 1

The correct option is C $3\sqrt{2}$

$x^2+y^2-4x-2y-4=0$

$x^2-4x+4-4+y^2-2y+1-1-4=0$

$\Rightarrow (x-2{)}^2+(y-1{)}^2=9$

Center $=(2,1)$

radius $=3$ units

for the triangle to be right-angled, the longest side should be the diameter of the circle

(since angle in a semicircle is ${90}^{\circ }$)

Since the triangle is isosceles, let the sides be $a,a,b$

$\Rightarrow b=2r=b$ units

$a^2+a^2=b^2$

$2a^2=b^2$

$\Rightarrow 2a^2=36$

$\Rightarrow a^2=18$

$\Rightarrow a=3\sqrt{2}$

$\therefore$ the sides of the triangle is $3\sqrt{2}$ units