The chord of contact of tangents from a point `P` to a circle passes through `Qdot` If `l_1a n dl_2` are the length of the tangents from `Pa n dQ` to the circle, then `P Q` is equal to

Options

A. `(l_(1)+l_(2))/(2)`
B. `(l_(1)-l_(2))/(2)`
C. `sqrt(l_(1)^(2)+l_(2)^(2))`
D. `2sqrt(l_(1)^(2)+l_(2)^(2))`

Answer

3

Explanation

Let `P(x_(1),y_(1))` and `Q (x_(2),y_(2))` be two points and `x^(2)+y^(20=a^(2)` be the given circle. Then the chord of contact of tangents drawn from P to the given circle is `x x_(1)+y y_(1)=a^(2)`.
It will pass through `Q(x_(2),y_(2))` if
`x_(1)x_(2)+y_(1)y_(2)=a^(2)` (1)
Now, `I_(1)=sqrt(x_(1)^(2)+y_(1)^(2)-a^(2))`
`I_(2)=sqrt(x_(1)^(2)+y_(1)^(2)-a^(2))`
and `PQ=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))`
`=sqrt((x_(2)^(2)-y_(2)^(2))+(x_(1)^(2)+y_(1)^(2))-2(x_(1)x_(2)+y_(1)y_(2)))`
`=sqrt([x_(1)^(2)+y_(1)^(2)-a^(2))+(x_(2)^(2)+y_(2)^(2)-a^(2)])` [Using Eq. (1) ]
`=sqrt((x_(1)^(2)+y_(1)^(2)-a^(2))+(x_(2)^(2)+y_(2)^(2)-a^(2))`
`=sqrt(l_(1)^(2)+l_(2)^(2))`