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
Correct Answer: C
Solution :
| [c] Let \[A\equiv (2,0)\] |
| Given lines are \[3x+5y=1\] ? (1) |
| and \[(2+c)x+5{{c}^{2}}y=1\] ? (2) |
| Multiplying equation (1) by \[{{c}^{2}}\] and subtracting (2) form it, |
| we get \[(3{{c}^{2}}-c-2)x={{c}^{2}}-1\] or \[x=\frac{{{c}^{2}}-1}{3{{c}^{2}}-c-2}\] |
| 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 \[\left( \frac{2}{5},-\frac{1}{25} \right)\] |
| Also, the circle passes through the point \[A(2,0)\] |
| \[\therefore \] radius of the circle |
| \[=\sqrt{{{\left( 2-\frac{2}{5} \right)}^{2}}+{{\left( 0+\frac{1}{25} \right)}^{2}}}\] |
| Thus, equation of the required circle is |
| \[={{\left( x-\frac{2}{5} \right)}^{2}}+{{\left( y+\frac{1}{25} \right)}^{2}}=\frac{64}{25}+\frac{1}{625}\] |
| or \[25({{x}^{2}}+{{y}^{2}})-20x+2y-60=0\] |
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