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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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