Question

Let $ABC$ be the triangle with $AB=1,AC=3$ and $\angle BAC=\frac{\pi }{2}$. If a circle of radius $r>0$ touches the sides $AB,AC$ and also touches internally the circumcircle of the triangle $ABC,$ then the value of $r$ is

• A. 0.8400
• B. 0.840
• C. .84
• D. 0.84

Answer 1

Answer: (A), (B), (C), (D)

Let $A$ be the origin, $B$ on x-axis, $C$ on y-axis as shown below

$\therefore$ Equation of circumcirle is
${(x-\frac{1}{2})}^2+{(y-\frac{3}{2})}^2={\left(\frac{1}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2=\frac{5}{2}\cdots \left(1\right)$
Required circle touches $AB$ and $AC,$ have radius $r$
$\therefore$ Equation be $(x-r{)}^2+(y-r{)}^2=r^2\cdots \left(2\right)$
If circle in equation $\left(2\right)$ touches circumcirle internally, we have
$d_{c_1{c}_2}=|r_1-r_2|$
$⟹{(\frac{1}{2}-r)}^2{(\frac{3}{2}-r)}^2={(\sqrt{\frac{5}{3}}-r)}^2$
$⟹\frac{1}{4}+r^2-r+\frac{9}{4}+r^2-3r={(\sqrt{\frac{5}{3}}-r)}^2\text{or}{(r-\sqrt{\frac{5}{3}})}^2$
$⟹2r^2-4r+\frac{5}{2}=\frac{5}{2}+r^2-\sqrt{10}r$
$⟹r=0\text{or}4-\sqrt{10}$
$⟹r=0.837$
$⟹r=0.84$ (on rounding off)