JEE Main 2022 — Mathematics Coordinate Geometry

Let the mirror image of a circle ${c}_{1}:{x}^{2}+{y}^{2}-2x-6y+\alpha =0$c1​:x2+y2−2x−6y+α=0 in line $y=x+1$y=x+1 be ${c}_{2}:5{x}^{2}+5{y}^{2}+10gx$c2​:5x2+5y2+10gx $+10fy+38=0$+10fy+38=0. If $r$r is the radius of circle ${c}_{2}$c2​, then $\alpha +6{r}^{2}$α+6r2 is equal to ______

Now Image of centre ${c}_{1}\equiv (1,3)$c1​≡(1,3) in $x-y+1=0$x−y+1=0 is given by

$$\frac{{x}_{1}-1}{1}=\frac{{y}_{1}-3}{-1}=\frac{-2(1-3+1)}{{1}^{2}+{1}^{2}}$$1x1​−1​=−1y1​−3​=12+12−2(1−3+1)​

$$\Rightarrow {x}_{1}=2,{y}_{1}=2$$⇒x1​=2,y1​=2

So, centre of circle ${c}_{2}\equiv (2,2)$c2​≡(2,2)

Now radius of circle ${c}_{2}$c2​ will be $\sqrt{4+4-\frac{38}{5}}=\sqrt{\frac{2}{5}}=r$4+4−538​​=52​​=r

Now $(\text{radius of}{{c}_{1}}^{2})=$(radius ofc1​2)=${(\text{radius of}{c}_{2})}^{2}$(radius ofc2​)2 as they are image to one another,

$$\Rightarrow \sqrt{{1}^{2}+{3}^{2}-\alpha }=\sqrt{\frac{2}{5}}$$⇒12+32−α​=52​​

$$\Rightarrow 10-\alpha =\frac{2}{5}\Rightarrow \alpha =\frac{48}{5}$$⇒10−α=52​⇒α=548​

Now putting the value of \alpha &r in $\alpha +6{r}^{2}$α+6r2 we get,

$$\alpha +6{r}^{2}=\frac{48}{5}+\frac{12}{5}=12$$α+6r2=548​+512​=12