Let a circle passing through (2,0) have its c | vedclass

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(A) Given lines are $3x + 5y = 1$ and $(2+c)x + 5c^2y = 1$.From the first line,$y = \frac{1-3x}{5}$. Substituting this into the second line:$(2+c)x +

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

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(A) Given lines are $3x + 5y = 1$ and $(2+c)x + 5c^2y = 1$.From the first line,$y = \frac{1-3x}{5}$. Substituting this into the second line:$(2+c)x + 5c^2(\frac{1-3x}{5}) = 1$$(2+c)x + c^2(1-3x) = 1$$x(2+c-3c^2) = 1-c^2$$x_c = \frac{1-c^2}{2+c-3c^2} = \frac{(1-c)(1+c)}{(1-c)(2+3c)} = \frac{1+c}{2+3c}$.$h = \lim_{c 	o 1} x_c = \frac{1+1}{2+3(1)} = \frac{2}{5}$.Now,$y_c = \frac{1-3x_c}{5} = \frac{1 - 3(\frac{1+c}{2+3c})}{5} = \frac{2+3c-3-3c}{5(2+3c)} = \frac{-1}{5(2+3c)}$.$k = \lim_{c 	o 1} y_c = \frac{-1}{5(2+3)} = -\frac{1}{25}$.The centre is $(h, k) = (\frac{2}{5}, -\frac{1}{25})$.The circle passes through $(2, 0)$,so the radius squared $r^2 = (2 - \frac{2}{5})^2 + (0 - (-\frac{1}{25}))^2 = (\frac{8}{5})^2 + (\frac{1}{25})^2 = \frac{64}{25} + \frac{1}{625} = \frac{1600+1}{625} = \frac{1601}{625}$.The equation is $(x - \frac{2}{5})^2 + (y + \frac{1}{25})^2 = \frac{1601}{625}$.$x^2 - \frac{4}{5}x + \frac{4}{25} + y^2 + \frac{2}{25}y + \frac{1}{625} = \frac{1601}{625}$.Multiplying by $625$: $625x^2 - 500x + 100 + 625y^2 + 50y + 1 = 1601$.$625x^2 + 625y^2 - 500x + 50y - 1500 = 0$.Dividing by $25$: $25x^2 + 25y^2 - 20x + 2y - 60 = 0$.

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 $(2+c)x + 5c^2y = 1$. If $h = \lim_{c \to 1} x_c$ and $k = \lim_{c \to 1} y_c$,then the equation of the circle is:

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