Question

The equation of the pair of straight lines joining the point of intersection of the curve $x^2+y^2=4$ and $y-x=2$ to the origin, is:

• A. $x^2+y^2={(y-x)}^2$
• B. $x^2+y^2+{(y-x)}^2=0$
• C. $x^2+y^2=4{(y-x)}^2$
• D. $x^2+y^2+4{(y-x)}^2=0$

Answer 1

The correct option is A

$x^2+y^2={(y-x)}^2$

Explanation for the correct option:

Finding the point of intersection,

Given $x^2+y^2=4$ and $y-x=2$ :

$$\begin{gathered} y-x=2 \\ \Rightarrow y=x+2 \\ \Rightarrow x^2+{(x+2)}^2=4\left[\text{Substituting}y\text{in}{x}^2+y^2=4\right] \\ \Rightarrow 2x^2+4x=0 \\ \Rightarrow x^2+2x=0 \\ \Rightarrow x(x+2)=0 \\ \Rightarrow x=0or-2 \\ \therefore y=2or0 \end{gathered}$$

Hence, the points of intersection are $A(0,2)\&B(-2,0)$.

The coordinates of origin is $(0,0)$.

Slope of $OA=\frac{0}{2}=0$

Slope of $OB=\frac{2}{0}$

Equation of the line $OA:$

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-0=0(x-0) \\ \Rightarrow y=0 \end{gathered}$$

Equation of the line $OB:$

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-0=\frac{2}{0}(x-0) \\ \Rightarrow 2x=0 \\ \Rightarrow x=0 \end{gathered}$$

So, the equation of the pair of straight lines is $xy=0$.

$\Rightarrow x^2+y^2={(y-x)}^2$

Hence, the correct answer is Option (A).