NIMCET 2018 Question Paper with solutions

On comparing to the standard equation of circle with $x^2+y^2$ + 2gx + 2fy + $C_1$ = 0 we get , g = - a , f = 0 , $C_1$ = $c^2$ Here, centre $C_1$ = (-g , -f) = (a , 0) and radius $(r_1)$ = $\sqrt{g^2+f^2-C_1}$ = $\sqrt{a^2-c^2}$ Equation of another circle is $x^2+y^2$ - 2by + $x^2$ = 0 ... (ii) Here, centre $C_2$ = (0 , b) and radius $(r_2)$ = $\sqrt{b^2-c^2}$ As we know that, two circles touch each other externally, if $C_1C_2$ = $r_1+r_2$ ⇒ $\sqrt{a^2+b^2}$ = $\sqrt{a^2-c^2}+\sqrt{b^2-c^2}$ Squaring both sides, we get ⇒ $a^2+b^2$ = $a^2-c^2+b^2-c^2$ + 2 $\sqrt{a^2-b^2}$ $\sqrt{b^2-c^2}$ ⇒ $2x^2$ = 2 $\sqrt{a^2-b^2}$ $\sqrt{b^2-c^2}$ ⇒ $c^2$ = $\sqrt{a^2-b^2}$ $\sqrt{b^2-c^2}$ Again squaring both sides, we get ⇒ $x^4$ = $(a^2-c^2)(b^2-c^2)$ ⇒ $c^4$ = $a^2b^2-a^2c^2-b^2c^2+c^4$ ⇒ $a^2c^2+b^2c^2$ = $a^2b^2$ ... (ii) Divide Eq. (iii) by $a^2b^2c^2$ , we get ⇒ $\frac{1}{b^2}+\frac{1}{a^2}$ = $\frac{1}{c^2}$ ⇒ $\frac{1}{c^2}$ = $\frac{1}{a^2}+\frac{1}{b^2}$