The lengths of the tangents from the points A and B to a circle are `l_(1)` and `l_(2)` respectively. If points are conjugate with respect to the circle, then `AB^(2)=`

Options

A. `l_(1)+l_(2)`
B. `l_(1)^(2)+l_(2)^(2)`
C. `|l_(1)^(2)-l_(2)^(2)|`
D. none of these

Answer

B

Explanation

Let the circle be `x^(2)+y^(2)=a^(2)`, and let `A(x_(1), y_(1)), B(x_(2), y_(2))` be the given points. Then,
`l_(1)^(2)-x_(1)^(2)+y_(1)^(2)-a^(2), l_(2)^(2)=x_(2)^(2)+y_(2)^(2)-a^(2)`
The polar of point A `(x_(1), y_(1))` with respect to `x^(2)+y^(2)=a^(2)` is `x x_(1)+y y_(1)=a^(2) " " ...(i)`
Since A and B are conjugate points. Therefore, `B(x_(2), y_(2))` lies on (i) i.e. `x_(1)x_(2)+y_(1)y_(2)=a^(2)`
Now,
`AB^(2)=(x_(2)-x_(2))^(2)+(y_(2)-y_(1))^(2)`
`rArr AB^(2)=(x_(1)^(2)+y_(1)^(2))+(x_(2)^(2)+y_(2)^(2))-2(x_(1)x_(2)+y_(1)y_(2))`
`rArr AB^(2)=(x_(1)^(2)-y_(1)^(2))+(x_(2)^(2)+y_(2)^(2))-2a^(2) ` [Using (i)]
`rArr AB^(2)=(x_(1)^(2)+y_(1)^(2)-a^(2))+(x_(2)^(2)+y_(2)^(2)-a^(2))=l_(1)^(2)+l_(2)^(2)`