If the straight line y = mx + c touches the c | vedclass

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(A) The equation of the circle is ${x^2} + {y^2} - 4y = 0$. Rewriting it as ${x^2} + {(y - 2)^2} = 4$,we identify the center as $(0, 2)$ and the radiu

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$2(1 \pm \sqrt{1 + {m^2}})$

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(A) The equation of the circle is ${x^2} + {y^2} - 4y = 0$. Rewriting it as ${x^2} + {(y - 2)^2} = 4$,we identify the center as $(0, 2)$ and the radius $r = 2$. For the line $mx - y + c = 0$ to be tangent to the circle,the perpendicular distance from the center $(0, 2)$ to the line must equal the radius $r$. $\frac{|m(0) - 2 + c|}{\sqrt{m^2 + (-1)^2}} = 2$ $|c - 2| = 2\sqrt{1 + m^2}$ $c - 2 = \pm 2\sqrt{1 + m^2}$ $c = 2 \pm 2\sqrt{1 + m^2} = 2(1 \pm \sqrt{1 + m^2})$.

If the straight line $y = mx + c$ touches the circle ${x^2} + {y^2} - 4y = 0$,then the value of $c$ will be

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