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

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(C) The equation of the circle is $x^2 + y^2 - 2x - 4y + 3 = 0$.The equation of the tangent to the circle $x^2 + y^2 + 2gx + 2fy + c' = 0$ at the poin

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$5$

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(C) The equation of the circle is $x^2 + y^2 - 2x - 4y + 3 = 0$.The equation of the tangent to the circle $x^2 + y^2 + 2gx + 2fy + c' = 0$ at the point $(x_1, y_1)$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c' = 0$.Comparing the given circle with the standard form,we have $g = -1$,$f = -2$,and $c' = 3$.Substituting the point $(2, 3)$ into the tangent formula:$x(2) + y(3) - 1(x + 2) - 2(y + 3) + 3 = 0$$2x + 3y - x - 2 - 2y - 6 + 3 = 0$$x + y - 5 = 0$Rearranging into the slope-intercept form $y = mx + c$:$y = -x + 5$Comparing this with $y = mx + c$,we get $c = 5$.

If the straight line $y = mx + c$ touches the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the point $(2, 3)$,then $c =$

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