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

:   Given Equation   $x^2+y^2-x-1=0--\left(1\right)andax+by=1--\left(2\right)$

Homogenizing (1) with (2)

$\begin{array}{l}{x}^2+y^2-x\left(1\right)-y\left(1\right)-1{\left(1\right)}^2=0\\ \Rightarrow x^2+y^2-x\left(ax+by\right)-y\left(ax+by\right)-1{\left(ax+by\right)}^2=0\\ \Rightarrow x^2+y^2-a{x}^2-bxy-axy-b{y}^2-\left(a{x}^2+b{y}^2+2abxy\right)=0\\ Thelinesrepresent\perp rlines\\ \therefore co.eff{x}^2+co.eff{y}^2=0\\ \Rightarrow 1+1-a-b-a^2-b^2=0\\ \Rightarrow a^2+b^2+a+b-2=0\\ Thelocusof\left(a,b\right)is{x}^2+y^2+x+y-2=0\\ r=\sqrt{\frac{1}{4}+\frac{1}{4}+2}=\sqrt{\frac{10}{4}}=\sqrt{\frac{5}{2}}\end{array}$