NIMCET 2012 Question Paper with solutions

Let ${\mathit{S}}_1$ = ${\mathit{x}}^2+{\mathit{y}}^2$ 2 +x + 2ky + 6 = 0
Here ${\mathit{g}}_1$ = 1 , ${\mathit{f}}_1$ = k , ${\mathit{C}}_1$ = 6 , Centre → (- 1 , k)
and ${\mathit{S}}_2$ = ${\mathit{x}}^2+{\mathit{y}}^2$ + 2ky + k = 0
Here, ${\mathit{g}}_2$ = 0 , ${\mathit{f}}_2$ = k and ${\mathit{C}}_2$ = k , Centre → (0 , - k)
If two circles intersect orthogonally, then
${\left(\mathrm{Distance}\mathrm{between}\mathrm{two}\mathrm{centres}\right)}^2$ = ${\left(\mathrm{Radius}\mathrm{of}\mathrm{circle}{\mathit{S}}_1\right)}^2$ + ${\left(\mathrm{Radius}\mathrm{of}\mathrm{circle}{\mathit{S}}_2\right)}^2$
${\left(-1-0\right)}^2+{\left(-\mathit{k}+\mathit{k}\right)}^2$ = ${\left(\sqrt{1+{\mathit{k}}^2-6}\right)}^2$ + ${\left(\sqrt{0+{\mathit{k}}^2-\mathit{k}}\right)}^2$
⇒ 1 + 0 = $\left({\mathit{k}}^2-5\right)$ + $\left({\mathit{k}}^2-\mathit{k}\right)$
⇒ 2${\mathit{k}}^2$ - k - 6 = 0
⇒ $2{\mathit{k}}^2$ - 4k + 3k - 6 = 0
⇒ 2k (k - 2) + 3 (k - 2) = 0
⇒ (k - 2) (2k + 3) = 0
∴ k = $-\frac{3}{2}$ or 2