Conic Sections Question 199
Question: The equation of a circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of the lines $ 3x+5y=1 $ and $ (2+c)x+5c^{2}y=1 $ as c tends to 1, is
Options:
A) $ 25(x^{2}+y^{2})+20x+2y-60=0 $
B) $ 25(x^{2}+y^{2})-20x+2y+60=0 $
C) $ 25(x^{2}+y^{2})-20x+2y-60=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ A\equiv (2,0) $ Given lines are $ 3x+5y=1 $
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(1) and $ (2+c)x+5c^{2}y=1 $
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(2) Multiplying equation (1) by $ c^{2} $ and subtracting (2) form it, we get $ (3c^{2}-c-2)x=c^{2}-1 $ or $ x=\frac{c^{2}-1}{3c^{2}-c-2} $ Now, $ \underset{c\to 1}{\mathop{\lim }}x=\underset{c\to 1}{\mathop{\lim }}\frac{(c-1)(c+1)}{(c-1)(3c+2)}=\underset{c\to 1}{\mathop{\lim }}\frac{c+1}{3c+2}=\frac{2}{5} $
$ \therefore $ X coordinate of centre $ =\frac{2}{5} $ From (1), when $ x=\frac{2}{5},y=-\frac{1}{25} $
Hence the centre of the circle is $ ( \frac{2}{5},-\frac{1}{25} ) $ Also, the circle passes through the point $ A(2,0) $
$ \therefore $ radius of the circle $ =\sqrt{{{( 2-\frac{2}{5} )}^{2}}+{{( 0+\frac{1}{25} )}^{2}}} $ Thus, equation of the required circle is $ ={{( x-\frac{2}{5} )}^{2}}+{{( y+\frac{1}{25} )}^{2}}=\frac{64}{25}+\frac{1}{625} $ or $ 25(x^{2}+y^{2})-20x+2y-60=0 $
BETA
