If the line x+2y=k intersects the curve x^2-x | vedclass

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(C) Given the curve equation: $x^2-xy+y^2+3x+3y-2=0$ $(i)$ and the line equation: $x+2y=k$,which implies $\frac{x+2y}{k}=1$. To find the condition for

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$2k^2+9k-10=0$

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(C) Given the curve equation: $x^2-xy+y^2+3x+3y-2=0$ $(i)$ and the line equation: $x+2y=k$,which implies $\frac{x+2y}{k}=1$. To find the condition for $\angle AOB=90^{\circ}$,we homogenize the curve equation using the line equation: $x^2-xy+y^2+(3x+3y)\left(\frac{x+2y}{k}\right)-2\left(\frac{x+2y}{k}\right)^2=0$. Multiplying by $k^2$: $k^2x^2-k^2xy+k^2y^2+3k(x^2+2xy+xy+2y^2)-2(x^2+4xy+4y^2)=0$. Expanding and grouping terms: $x^2(k^2+3k-2) + xy(-k^2+9k-8) + y^2(k^2+6k-8) = 0$. For $\angle AOB=90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(k^2+3k-2) + (k^2+6k-8) = 0$. $2k^2+9k-10=0$.

If the line $x+2y=k$ intersects the curve $x^2-xy+y^2+3x+3y-2=0$ at two points $A$ and $B$ and if $O$ is the origin,then the condition for $\angle AOB=90^{\circ}$ is

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