Question

The equation of pair of lines joining the origin to the point of intersection of $x^2+y^2=9$ and $x+y=3$ is:

• A. ${(x^2+-x)}^2=9$
• B. $xy=0$
• C. ${(3+y)}^2+y^2=9$
• D. ${(x-y)}^2=9$

Answer 1

The correct option is B

$xy=0$

Finding the point of intersection of $x^2+y^2=9$ and $x+y=3$:

$$\begin{gathered} x+y=3 \\ \Rightarrow y=3-x \\ \therefore x^2+{(3-x)}^2=9\left[\text{Substituting}yin{x}^2+y^2=9\right] \\ \Rightarrow 2x^2-6x+9=9 \\ \Rightarrow 2x^2-6x=0 \\ \Rightarrow x^2-3x=0 \\ \Rightarrow x(x-3)=0 \\ \Rightarrow x=0or3 \\ \therefore y=3or0 \end{gathered}$$

Therefore, the point of intersection are $(0,3)and(3,0)$.

We have to find the equation of the pair of lines that joins the origin to the points $(0,3)and(3,0)$.

We know that, $(0,3)$ lies on the y-axis while $(3,0)$ lies on the x-axis. Hence, the equation of $(0,3)$ is $x=0$and the equation of the point $(3,0)$ is $y=0$.

Therefore, the equation of the pair of lines is: $xy=0$.

Hence, the correct answer is Option (B).