Question

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 $(2+\alpha )x+5{\alpha }^{2}y=1$ as $\alpha$ tends to $1$ is

• A. $25(x^2+y^2)+20x-2y-140=0$
• B. $25(x^2+y^2)-20x+2y-60=0$
• C. $9(x^2+y^2)-20x+2y+4=0$
• D. $9(x^2+y^2)-2x-20y+4=0$

Answer 1

Answer: (B)

$25(x^2+y^2)-20x+2y-60=0$

Solving the given equation of the lines $3x+5y=1$ and $(2+\alpha )x+{5\alpha }^{2}y=1$
We get $x=\frac{\alpha +1}{3\alpha +2}=\frac{1+1}{3+2}=\frac{2}{5}$ as $\alpha \to 1$
Thus, $3\times \frac{2}{5}+5y=1\Rightarrow 5y=1-\frac{6}{5}=\frac{-1}{5}\Rightarrow y=\frac{-1}{25}$
Henve the center of the circle is $(\frac{2}{5},-\frac{1}{25})$
If r be the radius of the circle, then its equation is
${(2-\frac{2}{5})}^2+{(0+\frac{1}{25})}^2=r^2\Rightarrow r^2=\frac{1601}{625}$
The equation of the required circle is
${(x-\frac{2}{5})}^2+{(y+\frac{1}{25})}^2=\frac{1601}{625}\Rightarrow 25(x^2+y^2)-20x+2y-60=0$