If the lines joining the origin to the points | vedclass

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(C) The given curve is $x^2+y^2-2x-4y+2=0$ and the line is $x+y=k$,which implies $\frac{x+y}{k}=1$. To find the equation of the lines joining the orig

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$3$

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(C) The given curve is $x^2+y^2-2x-4y+2=0$ and the line is $x+y=k$,which implies $\frac{x+y}{k}=1$. To find the equation of the lines joining the origin to the intersection points,we homogenize the curve equation using the line equation: $x^2+y^2-2(x+2y)(\frac{x+y}{k})+2(\frac{x+y}{k})^2=0$. Multiplying by $k^2$,we get: $k^2(x^2+y^2)-2k(x+2y)(x+y)+2(x+y)^2=0$. Expanding the terms: $k^2x^2+k^2y^2-2k(x^2+3xy+2y^2)+2(x^2+y^2+2xy)=0$. Grouping the coefficients of $x^2$ and $y^2$: $x^2(k^2-2k+2)+y^2(k^2-4k+2)+xy(-6k+4)=0$. Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(k^2-2k+2)+(k^2-4k+2)=0$. $2k^2-6k+4=0 \implies k^2-3k+2=0$. Factoring the quadratic equation: $(k-1)(k-2)=0$,so $k=1$ or $k=2$. The sum of all possible values of $k$ is $1+2=3$.

If the lines joining the origin to the points of intersection of the line $x+y=k$ and the curve $x^2+y^2-2x-4y+2=0$ are at right angles,then the sum of all the possible values of $k$ is

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