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વર્તુળ $C_1: x^2+y^2-4 x-2 y=\alpha-5$ ધ્યાને લો.ધારોકે તેનુ રેખા $y=2 x+1$ પરનું આરસી પ્રતિબિંબ અન્ય વર્તુળ $C_2: 5 x^2+5 y^2-10 f x-10 g y+36=0$ છે. ધારોકે $r$ એ $C_2$ ની ત્રિજયા છે. તો $\alpha+r=.......$
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Solution
$x^2+y^2-4 x-2 y+5-\alpha=0,$
$C_1(2,1) r_1=\sqrt{\alpha}$
$2 x-y+1=0$
Image of $(2,1)$
$\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-2(4-1+1)}{5}$
$\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5}$
$x=2-\frac{16}{5}=\frac{-6}{5}, y=1+\frac{8}{5}=\frac{13}{5}$
$x^2+y^2-2 f x-2 g y+\frac{36}{5}=0$
$C_2(f, g)$
$r _2=\sqrt{ f ^2+ g ^2-\frac{36}{5}}$
$\alpha= f ^2+ g ^2-\frac{36}{5}$
$\therefore f =-\frac{6}{5}, g =\frac{13}{5}$
$\alpha=\frac{36}{25}+\frac{169}{25}-\frac{36}{5}$
$=\frac{36+169-180}{25} \Rightarrow \alpha=1 \Rightarrow r =1$
$\therefore \alpha+ r =2$
