The equation of the pair of lines joining the origin to the points of intersection of $x^2+y^2=9$ and…
The equation of the pair of lines joining the origin to the points of intersection of $x^2+y^2=9$ and $x+y=3$, is
- $x^2+(3-y)^2=9$
- $(3+y)^2+y^2=9$
- $x^2 - y^2 = 9$
- $xy = 0$
Solution
Given, $x^2+y^2=9$ and $x+y=3$ Equation of pair of lines.
$\begin{aligned}
& x^2+y^2=9\left(\frac{x+y}{3}\right)^2 \\
& \Rightarrow \quad x^2+y^2=(x+y)^2
\end{aligned}$
$\begin{aligned} & \Rightarrow \quad x^2+y^2=x^2+y^2+2 x y \\ & \therefore \quad x y=0\end{aligned}$
Asked in: AP EAMCET 2016
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