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(C) The equation of the line is $ax+by=1$,which can be written as $ax+by=1$. The equation of the curve is $x^2+y^2-x-y-1=0$.By homogenizing the equati

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$\sqrt{5/2}$

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(C) The equation of the line is $ax+by=1$,which can be written as $ax+by=1$. The equation of the curve is $x^2+y^2-x-y-1=0$.By homogenizing the equation of the curve using the line equation,we get:$x^2+y^2-(x+y)(ax+by)-(ax+by)^2=0$$x^2+y^2-(ax^2+bxy+axy+by^2)-(a^2x^2+b^2y^2+2abxy)=0$$x^2(1-a-a^2)+xy(-a-b-2ab)+y^2(1-b-b^2)=0$Since the pair of lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(1-a-a^2)+(1-b-b^2)=0$$a^2+b^2+a+b-2=0$This represents a circle in the $(a, b)$ plane. Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we have $g=1/2$,$f=1/2$,and $c=-2$.The radius $r$ is given by $\sqrt{g^2+f^2-c}$:$r = \sqrt{(1/2)^2+(1/2)^2-(-2)}$$r = \sqrt{1/4+1/4+2} = \sqrt{1/2+2} = \sqrt{5/2}$.

If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

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