If the segments of the straight lines $x+y=6$ and $x+2 y=4$ are two diameters of a circle passing through…

If the segments of the straight lines $x+y=6$ and $x+2 y=4$ are two diameters of a circle passing through $(6,2)$, then the equation of that circle is
  1. $x^2+y^2-2 x-4 y-20=0$
  2. $x^2+y^2+6 x-4 y-68=0$
  3. $x^2+y^2-16 x+4 y+48=0$
  4. $x^2+y^2+2 x-10 y-32=0$

Solution

$L_1 \equiv x+y=6$ and $L_2 \equiv x+2 y=4$
Point of intersection of lines $L_1$ and $L_2$ is centre $C$ of the circle. Solving $L_1$ and $L_2$ : $\because \quad x+y=6 \Rightarrow x=6-y$ Then, $x+2 y=4$ $\begin{array}{ll}\Rightarrow & 6-y+2 y=4 \\ \Rightarrow & y=-2 \text { and } x=8\end{array}$ Thus, $C: \equiv(8,-2)$. CP is the radius. $C P=\sqrt{(8-6)^2+(2+2)^2}=\sqrt{4+16}=\sqrt{20}$ Equation of circle $:(x-8)^2+(y+2)^2=(\sqrt{20})^2$ $\Rightarrow \quad x^2+y^2-16 x+4 y+48=0$

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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