The equation of the circle whose centre is (1 | vedclass

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(A) The radius $r$ of the circle is the perpendicular distance from the center $(1, -3)$ to the line $2x - y - 4 = 0$.$r = \left| \frac{2(1) - (-3) -

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$5x^2 + 5y^2 - 10x + 30y + 49 = 0$

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(A) The radius $r$ of the circle is the perpendicular distance from the center $(1, -3)$ to the line $2x - y - 4 = 0$.$r = \left| \frac{2(1) - (-3) - 4}{\sqrt{2^2 + (-1)^2}} \right| = \left| \frac{2 + 3 - 4}{\sqrt{5}} \right| = \frac{1}{\sqrt{5}}$.The equation of the circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 1)^2 + (y + 3)^2 = \left( \frac{1}{\sqrt{5}} \right)^2$.$(x^2 - 2x + 1) + (y^2 + 6y + 9) = \frac{1}{5}$.$x^2 + y^2 - 2x + 6y + 10 = \frac{1}{5}$.Multiplying by $5$,we get $5x^2 + 5y^2 - 10x + 30y + 50 = 1$.$5x^2 + 5y^2 - 10x + 30y + 49 = 0$.

The equation of the circle whose centre is $(1, -3)$ and which touches the line $2x - y - 4 = 0$ is

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