10-1.Circle and System of Circles
normal

माना रेखा $y=x+1$ में, वृत्त $c_1: x^2+y^2-2 x-6 y+$ $\alpha=0$ का दर्पण प्रतिबंब $c_2: 5 x^2+5 y^2+10 gx +$ $10 fy +38=0$ है। यदि वृत्त $c _2$ की त्रिज्या $r$ है, तो $\alpha+6 r^2$ बराबर है $...........।$

A

$13$

B

$11$

C

$12$

D

$10$

(JEE MAIN-2022)

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 $\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}= r$

$\left(\text { radius of } c _{1}\right)^{2}=\left(\text { radius of } c _{2}\right)^{2}$

$10-\alpha=\frac{2}{5} \Rightarrow \alpha=\frac{48}{5}$

$\therefore \alpha+6 r ^{2}=\frac{48}{5}+\frac{12}{5}=12$

Standard 11
Mathematics

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