From an arbitrary point `P` on the circle `x^2+y^2=9` , tangents are drawn to the circle `x^2+y^2=1` , which meet `x^2+y^2=9` at `A` and `B` . The locus of the point of intersection of tangents at `A` and `B` to the circle `x^2+y^2=9` is
(a) `x^2+y^2=((27)/7)^2` (b) `x^2-y^2((27)/7)^2` `y^2-x^2=((27)/7)^2` (d) none of these

Options

A. `x^(2)+y^(2)=(27//7)^(2)`
B. `x^(2)-y^(2)=(27//7)^(2)`
C. `y^(2)-x^(2)=(27//7)^(2)`
D. none of these

Answer

1

Explanation

Since `Delta POQ` and `Delta AOQ` are congruent,
`/_POQ= /_ QOA= theta`

`cos theta =(1)/(2), ` since `/_POR =pi`
`:. /_ AOR =pi-2 theta`
Now, in triangle `AOR, /_AOR =pi -2theta` and `AO =3` units
Therefore,
`cos (pi-2theta)=(OA)/(OR)=(3)/(sqrt(h^(2)+k^(2)))`
or `sqrt(h^(2)+k^(2))=(27)/(7)`
or `x^(2)+y^(2)=((27)/(7))^(2)`