Let a circle passing through (2,0) have its centre at the point (h, k). Let be the point of intersection of the lines 3x + 5y = 1 and . If and , then the equation of the circle is : [2024]

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Solution

Q.
Let a circle passing through (2,0) have its centre at the point (h, k). Let $(x_{c}, y_{c})$ be the point of intersection of the lines 3x + 5y = 1 and $(2 + c) x + 5 c^{2} y = 1$ . If $h = \lim_{c \to 1} x_{c}$ and $k = \lim_{c \to 1} y_{c}$ , then the equation of the circle is : [2024]

1 $5 x^{2} + 5 y^{2} - 4 x + 2 y - 12 = 0$

2 $5 x^{2} + 5 y^{2} - 4 x - 2 y - 12 = 0$

3 $25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0$

4 $25 x^{2} + 25 y^{2} - 2 x + 2 y - 60 = 0$

Ans.
(3)

We have, 3x + 5y = 1 ... (i)

$(2 + c) x + 5 c^{2} y = 1$ ... (ii)

Multiplying (i) by $c^{2}$ and subtracting it from (ii), we get

$(2 + c - 3 c^{2}) x = 1 - c^{2}$

$\Rightarrow x_{c} = \frac{1 - c^{2}}{2 + c - 3 c^{2}} = \frac{(1 - c) (1 + c)}{(1 - c) (3 c + 2)} = \frac{c + 1}{3 c + 2}$

$∴ y = \frac{1 - 3 x}{5} = \frac{1 - 3 (\frac{c + 1}{3 c + 2})}{5} = \frac{3 c + 2 - 3 c - 3}{5 (3 c + 2)}$

$y_{c} = \frac{- 1}{5 (3 c + 2)}$

Now, $h = \lim_{c \to 1} x_{c} = \lim_{c \to 1} (\frac{c + 1}{3 c + 2}) = \frac{1 + 1}{3 + 2} = \frac{2}{5}$

and $k = \lim_{c \to 1} y_{c} = \lim_{c \to 1} \frac{- 1}{5 (3 c + 2)} = \frac{- 1}{25}$

Equation of circle is

${(x - \frac{2}{5})}^{2} + {(y + \frac{1}{25})}^{2} = \frac{64}{25} + \frac{1}{625}$ [ $∵$ Circle passes through (2, 0)]

$\Rightarrow x^{2} + y^{2} - \frac{4}{5} x + \frac{2}{25} y + \frac{4}{25} + \frac{1}{625} = \frac{64}{25} + \frac{1}{625}$

$\Rightarrow 25 x^{2} + 25 y^{2} - 20 x + 2 y + 4 = 64$

$\Rightarrow 25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0$

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