Question

The tangent to a circle at P and Q intersect at T. AB is a diameter parallel to PQ. The lines joining P and Q with one A and B intersect the diameter perpendicular to AB at R and S. Then $\frac{RT}{ST}$ is

Answer 1

Let us take the circle $x^2+y^2=r^2$Let P be $(x_1,y_1)$ and Q be $(x_2,y_2)$Equation at tangent at P$x{x}_1+y{y}_1=r^2$at Q$x{x}_2+y{y}_2=r^2$For point at intersection$\frac{x}{-r^2{y}_1+r^2{y}_2}=\frac{y}{-r^2{x}_2+r^2{x}_1}=\frac{1}{x_1{y}_2-x_2{y}_1}T(x,y)=(\frac{r^2(y_2-y_1)}{x_1{y}_2-x_2{y}_1},\frac{r^2(x_1-x_2)}{x_1{y}_2-x_2{y}_1})$