Question

The locus of the point of intersection of perpendicular tangents to the circles $x^2+y^2=a^{2}and{x}^2+y^2=b^{2}is$

• A. x² + y² = a² + b²
• B. x² + y² = a²– b²
• C. (x + y)² = a² + b²
• D. x²– y² = a²– b²

Answer 1

The correct option is A

x² + y² = a² + b²

$\text{The equation of any tangent to}{x}^2+y^2=a^2\text{is}xcos\alpha +ysin\alpha =a....\left(1\right)\text{The equation of any tangent to}{x}^2+y^2=b^2,\text{perpendicular to}xcos\alpha +ysin\alpha =aisxsin\alpha -ycos\alpha =b\left(2\right)\text{Let P(h, k) be the point of intersection of (1) \& (2). Then}hcos\alpha +ksin\alpha =a,hsin\alpha -kcos\alpha =b\text{Squaring and adding}{h}^2+k^2=a^2+b^2\therefore \text{locus of (h, k) is}{x}^2+y^2=a^2+b^2$