Question

In a right angled triangle $BC=5,AB=4,AC=3$. Let $S$ be the circum circle. Let $S_1$ be the circle touching both sides $AB$ and $AC$ and circle $S$ internally. Let $S_2$ be the circle touching the sides $AB$ and $AC$ of $△ABC$ and touching the circle $S$ externally. If $r_1{r}_2$ are radii of circles $S_1$ and $S_2$ respectively then $r_1{r}_2=$

• A. 12
• B. 24
• C. 15
• D. 42

Answer 1

The correct option is B 24

Equation of $S=(x-\frac{a}{2}{)}^2+(y-\frac{b}{2}{)}^2=\frac{a^2+b^2}{4}$

$x^2+y^2-ax-by=0$

And circle with centre $(r,r)$ and radius $r$

Touches AB and AC its equation is

$(x-r{)}^2+(y-r{)}^2=r^2$

$x^2+y^2-2xr-2yr+r^2=0$

Both circle touches internally or externally

Using $\sqrt{(r-\frac{a}{2}{)}^2+(r-\frac{b}{2}{)}^2}=\sqrt{(\frac{a^2}{4}+\frac{b^2}{4})}\pm r$

$\Delta ABC=\frac{r_1{r}_2}{4}⟹r_1{r}_2=4\Delta ABC$

$ar\left(\Delta ABC\right)=\sqrt{6(6-5)(6-4)(6-3)}=\sqrt{6\times 1\times 2\times 3}=\sqrt{36}=6$

$\therefore r_1{r}_2=4\times 6=24$