NIMCET 2025 Paper

The original circle
${\mathit{x}}^2+{\mathit{y}}^2-6\mathit{x}-10\mathit{y}+30=0$
has centre

We reflect the centre across the line
$2\mathit{x}+\mathit{y}-2=0$
Reflection formula gives the image point ${\mathit{C}}^{\prime }\left({\mathit{x}}^{\prime },{\mathit{y}}^{\prime }\right)$ :
$\mathit{d}=\frac{2\left(3\right)+1\left(5\right)-2}{2^2+1^2}=\frac{9}{5}.$
Thus,
${\mathit{x}}^{\prime }=3-2\left(2\right)\left(\frac{9}{5}\right)=3-\frac{36}{5}=-\frac{21}{5}$
${\mathit{y}}^{\prime }=5-2\left(1\right)\left(\frac{9}{5}\right)=5-\frac{18}{5}=\frac{7}{5}$
So the image centre is
${\mathit{C}}^{\prime }=\left(-\frac{21}{5},\frac{7}{5}\right).$
Radius remains 2 .
Its equation is:
${\left(\mathit{x}-{\mathit{x}}^{\prime }\right)}^2+{\left(\mathit{y}-{\mathit{y}}^{\prime }\right)}^2=4.$

Expanding:
${\mathit{x}}^2+{\mathit{y}}^2+\frac{42}{5}\mathit{x}-\frac{14}{5}\mathit{y}+\left({\mathit{x}}^{\prime 2}+{\mathit{y}}^{\prime 2}-4\right)=0$
Compute constant:
${\mathit{x}}^{\prime 2}+{\mathit{y}}^{\prime 2}=\frac{441}{25}+\frac{49}{25}=\frac{490}{25}=\frac{98}{5}$
So:
$\mathit{\gamma }=\frac{98}{5}-4=\frac{98-20}{5}=\frac{78}{5}.$
Thus:
$\mathit{\alpha }+\mathit{\beta }+\mathit{\gamma }=\frac{42}{5}-\frac{14}{5}+\frac{78}{5}=\frac{106}{5}=21.2.$
$\left[\mathit{\alpha }+\mathit{\beta }+\mathit{\gamma }\right]=\lfloor 21.2\rfloor =21.$