NIMCET 2018 Question Paper with solutions

On comparing to the standard equation of circle with
${\mathit{x}}^2+{\mathit{y}}^2$ + 2gx + 2fy + ${\mathit{C}}_1$ = 0
we get , g = - a , f = 0 , ${\mathit{C}}_1$ = ${\mathit{c}}^2$
Here, centre ${\mathit{C}}_1$ = (-g , -f) = (a , 0)
and radius $\left({\mathit{r}}_1\right)$ = $\sqrt{{\mathit{g}}^2+{\mathit{f}}^2-{\mathit{C}}_1}$ = $\sqrt{{\mathit{a}}^2-{\mathit{c}}^2}$
Equation of another circle is
${\mathit{x}}^2+{\mathit{y}}^2$ - 2by + ${\mathit{x}}^2$ = 0 ... (ii)
Here, centre ${\mathit{C}}_2$ = (0 , b)
and radius $\left({\mathit{r}}_2\right)$ = $\sqrt{{\mathit{b}}^2-{\mathit{c}}^2}$
As we know that, two circles touch each other externally, if
${\mathit{C}}_1{\mathit{C}}_2$ = ${\mathit{r}}_1+{\mathit{r}}_2$
⇒ $\sqrt{{\mathit{a}}^2+{\mathit{b}}^2}$ = $\sqrt{{\mathit{a}}^2-{\mathit{c}}^2}+\sqrt{{\mathit{b}}^2-{\mathit{c}}^2}$
Squaring both sides, we get
⇒ ${\mathit{a}}^2+{\mathit{b}}^2$ = ${\mathit{a}}^2-{\mathit{c}}^2+{\mathit{b}}^2-{\mathit{c}}^2$ + 2 $\sqrt{{\mathit{a}}^2-{\mathit{b}}^2}$ $\sqrt{{\mathit{b}}^2-{\mathit{c}}^2}$
⇒ $2{\mathit{x}}^2$ = 2 $\sqrt{{\mathit{a}}^2-{\mathit{b}}^2}$ $\sqrt{{\mathit{b}}^2-{\mathit{c}}^2}$
⇒ ${\mathit{c}}^2$ = $\sqrt{{\mathit{a}}^2-{\mathit{b}}^2}$ $\sqrt{{\mathit{b}}^2-{\mathit{c}}^2}$
Again squaring both sides, we get
⇒ ${\mathit{x}}^4$ = $\left({\mathit{a}}^2-{\mathit{c}}^2\right)\left({\mathit{b}}^2-{\mathit{c}}^2\right)$
⇒ ${\mathit{c}}^4$ = ${\mathit{a}}^2{\mathit{b}}^2-{\mathit{a}}^2{\mathit{c}}^2-{\mathit{b}}^2{\mathit{c}}^2+{\mathit{c}}^4$
⇒ ${\mathit{a}}^2{\mathit{c}}^2+{\mathit{b}}^2{\mathit{c}}^2$ = ${\mathit{a}}^2{\mathit{b}}^2$ ... (ii)
Divide Eq. (iii) by ${\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2$ , we get
⇒ $\frac{1}{{\mathit{b}}^2}+\frac{1}{{\mathit{a}}^2}$ = $\frac{1}{{\mathit{c}}^2}$
⇒ $\frac{1}{{\mathit{c}}^2}$ = $\frac{1}{{\mathit{a}}^2}+\frac{1}{{\mathit{b}}^2}$