TS EAMCET 20-July-2022 Shift 1 Solved Paper

Line $x+2y=1$ cuts the $X$-axis at $A$ and $Y$-axis at $B$. Then, $A(1,0)$ and $B \left(0, \;\frac{1}{2}\right)$ A circle passes through $A$ and $B$ and origin. So, equation of circle is $x^2+y^2+2gx+2fy=0$ It passes through the point $A(1,0)$, $1+0+2g=0 \Rightarrow g=-\;\frac{1}{2}$ Also, passes through $\left(0, \;\frac{1}{2}\right)^2$$0+\;\frac{1}{4}+0+f=0 \Rightarrow f=-\;\frac{1}{4}$ So, equation of circle is $x^2+y^2-x-\frac{y}{2}=0$ So, tangent at $(0,0)$ is $\;x x_1+y y_1- \left( \;\frac{x+x_1}{2} \right) -\;\frac{1}{2} \left( \;\frac{y+y_1}{2} \right) =0$$\;0+0- \left( \;\frac{x+0}{2} \right) -\;\frac{1}{2} \left( \;\frac{y+0}{2} \right) =0$ $\Rightarrow\;\; 2x+y=0$ Perpendicular distance from $A(1,0)$ to tangent at origin $= \left| \;\frac{2+0}{\sqrt{5}} \right| =\;\frac{2}{\sqrt{5}}$. Perpendicular distance from $B \left(0, \;\frac{1}{2}\right)$ to tangent at origin $= \left| \;\frac{0+\frac{1}{2}}{\sqrt{5}} \right| =\;\frac{1}{2\sqrt{5}}$ Sum of perpendicular distances from $A$ and $B$ to tangent at origin $= \left(2+\;\frac{1}{2}\right) \;\frac{1}{\sqrt{5}} =\;\frac{\sqrt{5}}{2}$ Centre $(-y_1,-f) = \left(\;\frac{1}{2}, \;\frac{1}{4}\right)$ $\;\; \text{ Radius } = \sqrt{g^2+f^2-c} = \sqrt{\;\frac{1}{4}+\;\frac{1}{16}-0} =\;\frac{\sqrt{5}}{4}$$\;\; \text{ Diameter } = 2r =\;\frac{\sqrt{5}}{2}$ So, sum of perpendicular distances from $A$ and $B$ equal to the diameter of circle.