The straight line x+2y=1 cuts the X-axis at A | vedclass

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(B) The line is $x+2y=1$. The $X$-intercept is $A(1,0)$ and the $Y$-intercept is $B(0, 1/2)$.The circle passes through $(0,0), (1,0), (0, 1/2)$. The e

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equal to the diameter of the circle $S$

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(B) The line is $x+2y=1$. The $X$-intercept is $A(1,0)$ and the $Y$-intercept is $B(0, 1/2)$.The circle passes through $(0,0), (1,0), (0, 1/2)$. The equation of the circle is $x^2+y^2-x-\frac{1}{2}y=0$.The tangent at the origin $(0,0)$ is found by setting the linear terms to zero: $-x-\frac{1}{2}y=0$,which simplifies to $2x+y=0$.The perpendicular distance from $A(1,0)$ to $2x+y=0$ is $d_1 = \frac{|2(1)+0|}{\sqrt{2^2+1^2}} = \frac{2}{\sqrt{5}}$.The perpendicular distance from $B(0, 1/2)$ to $2x+y=0$ is $d_2 = \frac{|2(0)+1/2|}{\sqrt{2^2+1^2}} = \frac{1/2}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$.The sum of distances is $d_1+d_2 = \frac{2}{\sqrt{5}} + \frac{1}{2\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$.The radius of the circle $x^2+y^2-x-\frac{1}{2}y=0$ is $r = \sqrt{g^2+f^2-c} = \sqrt{(-1/2)^2+(-1/4)^2-0} = \sqrt{1/4+1/16} = \sqrt{5/16} = \frac{\sqrt{5}}{4}$.Thus,the sum of distances is $\frac{\sqrt{5}}{2} = 2 	imes \frac{\sqrt{5}}{4} = 2r$,which is the diameter of the circle $S$.

The straight line $x+2y=1$ cuts the $X$-axis at $A$ and the $Y$-axis at $B$. $A$ circle is drawn through $A, B$ and the origin $O(0,0)$. The sum of the perpendicular distances from $A$ and $B$ to the tangent drawn at the origin to the circle $S$ is:

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