The line x+2y-c=0 meets the curve x^2+y^2-3x- | vedclass

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(B) The equation of the circle is $x^2+y^2-3x-6y+3=0$. The line is $x+2y=c$,or $\frac{x+2y}{c}=1$. Homogenizing the circle equation using the line: $x

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-$15$

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(B) The equation of the circle is $x^2+y^2-3x-6y+3=0$. The line is $x+2y=c$,or $\frac{x+2y}{c}=1$. Homogenizing the circle equation using the line: $x^2+y^2-(3x+6y)(\frac{x+2y}{c}) + 3(\frac{x+2y}{c})^2 = 0$. Since $\angle POQ = \frac{\pi}{2}$,the sum of the coefficients of $x^2$ and $y^2$ must be $0$. Coefficient of $x^2$: $1 - \frac{3}{c} + \frac{3}{c^2} = 1 - \frac{3}{c} + \frac{12}{c^2}$. Coefficient of $y^2$: $1 - \frac{12}{c} + \frac{12}{c^2}$. Sum: $1 - \frac{3}{c} + \frac{3}{c^2} + 1 - \frac{12}{c} + \frac{12}{c^2} = 0$. $2 - \frac{15}{c} + \frac{15}{c^2} = 0$. Multiplying by $c^2$: $2c^2 - 15c + 15 = 0$. Thus,$2c^2 - 15c = -15$.

The line $x+2y-c=0$ meets the curve $x^2+y^2-3x-6y+3=0$ at two points $P$ and $Q$ and $\angle POQ = \frac{\pi}{2}$,where $O$ is the origin. Then $2c^2-15c =$

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