- Home
- Standard 11
- Mathematics
અહી વર્તુળ $c_{1}: x^{2}+y^{2}-2 x-$ $6 y+\alpha=0$ નું રેખા $y=x+1$ ની સાપેક્ષે પ્રતિબિંબ $c_{2}: 5 x^{2}+5 y^{2}+10 g x+10 f y +38=0$ છે. જો $r$ એ વર્તુળ $c _{2}$ ત્રિજ્યા હોય તો $\alpha+6 r^{2}$ ની કિમંત મેળવો.
$13$
$11$
$12$
$10$
Solution
Image of centre $c _{1} \equiv(1,3)$ in $x – y +1=0$ is given by
$\frac{x_{1}-1}{1}=\frac{y_{1}-3}{-1}=\frac{-2(1-3+1)}{1^{2}+1^{2}}$
$x_{1}=2, y_{1}=2$
$\therefore$ Centre of circle $c _{2} \equiv(2,2)$
$\therefore$ Equation of $c_{2}$ be $x^{2}+y^{2}-4 x-4 y+\frac{38}{5}=0$
Now radius of $c _{2}$ is $\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}= r$
$\left(\text { radius of } c _{1}\right)^{2}=\left(\text { radius of } c _{2}\right)^{2}$
$10-\alpha=\frac{2}{5} \Rightarrow \alpha=\frac{48}{5}$
$\therefore \alpha+6 r ^{2}=\frac{48}{5}+\frac{12}{5}=12$
