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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
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Verified Answer
The correct answer is:
$x^2+y^2-16 x+4 y+48=0$
$L_1 \equiv x+y=6$
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 $:(x-8)^2+(y+2)^2=(\sqrt{20})^2$
$\Rightarrow \quad x^2+y^2-16 x+4 y+48=0$
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 $:(x-8)^2+(y+2)^2=(\sqrt{20})^2$
$\Rightarrow \quad x^2+y^2-16 x+4 y+48=0$
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