Aspire Faculty ID #11977 · Topic: NIMCET 2025 · Just now
NIMCET 2025

The circle $x^2 + y^2+ \alpha x+ \beta y+ \gamma=0$ is the image of the circle $x^2 + y^2- 6x- 10y+ 30=0$ across the line 3x + y = 2. The value of $[\alpha+ \beta+ \gamma]$ is (where [.] represents the floor function.)

Solution

Given circle: 
$x^{2}+y^{2}-6x-10y+30=0$ 
Center and radius: $C(3,5), \quad r=2$ 

We reflect $C(3,5)$ across the line $3x + y - 2 = 0$. 
Compute reflection factor: $k = \dfrac{2(3\cdot 3 + 5 - 2)}{3^{2} + 1^{2}} = \dfrac{24}{10} = \dfrac{12}{5}$ 

Reflected center: 
$x' = 3 - 3k = 3 - \dfrac{36}{5} = -\dfrac{21}{5}$ 
$y' = 5 - k = 5 - \dfrac{12}{5} = \dfrac{13}{5}$ 
So new center is: 
$C'\left(-\dfrac{21}{5},\dfrac{13}{5}\right)$ 
The image circle is $x^{2}+y^{2}+\alpha x+\beta y+\gamma = 0$ 
Using center formula: $-\dfrac{\alpha}{2} = -\dfrac{21}{5} \Rightarrow \alpha = \dfrac{42}{5}$ 
$-\dfrac{\beta}{2} = \dfrac{13}{5} \Rightarrow \beta = -\dfrac{26}{5}$ 
Using radius $r=2$: $\left(-\dfrac{21}{5}\right)^{2} + \left(\dfrac{13}{5}\right)^{2} - \gamma = 4$ 
$\dfrac{610}{25} - \gamma = 4$ 
$\gamma = \dfrac{102}{5}$ 
Now compute: 
$\alpha + \beta + \gamma = \dfrac{42}{5} - \dfrac{26}{5} + \dfrac{102}{5} = \dfrac{118}{5} = 23.6$ 
Thus, $\boxed{23}$
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