The locus of the point of intersection of perpendicular tangents to the circles `x^(2)+y^(2)=a^(2)` and `x^(2)+y^(2)=b^(2)` , is

Options

A. `x^(2)+y^(2)=a^(2)-b^(2)`
B. `x^(2)+y^(2)=a^(2)+b^(2)`
C. `x^(2)+y^(2)=(a+b)^(2)`
D. none of these

Answer

B

Explanation

The equation of any tangent to `x^(2)+y^(2)=a^(2)` is
`x cos alpha + y sin alpha =a " " ...(i)`
The equation of the tangent to `x^(2)+y^(2)=b^(2)`, perpendicular to (i) is
`x sin alpha - y sin alpha = b " " ...(ii)`
Let (h, k) be the point of intersection of (i) and (ii). Then
`h cos alpha + k sin alpha = a " " ...(iii)`
and , ` h sin alpha- k cos alpha = b " " ...(iv)`
Squaring and adding (iii) and (iv), we get
`h^(2)+k^(2)=a^(2)+b^(2)`
Hence, the locus of (h, k) is `x^(2)+y^(2)=a^(2)+b^(2)`