A tangent at a point on the circle `x^2+y^2=a^2` intersects a concentric circle `C` at two points `Pa n dQ` . The tangents to the circle `X` at `Pa n dQ` meet at a point on the circle `x^2+y^2=b^2dot` Then the equation of the circle is `x^2+y^2=a b` `x^2+y^2=(a-b)^2` `x^2+y^2=(a+b)^2` `x^2+y^2=a^2+b^2`

Options

A. `x^(2)+y^(2)=ab`
B. `x^(2)+y^(2)=(a-b)^(2)`
C. `x^(2)+y^(2)=(a+b)^(2)`
D. `x^(2)+y^(2)=a^(2)=b^(2)`

Answer

1

Explanation

Chord of contact of point A w.r.t `x^(2)+y^(2)=r^(2)` is
`xb cos theta + y b sin theta =2` (1)
This must be a tangent to the circle `x^(2)+y^(2)=a^(2)`. Therefore,
`[(r^(2))/(sqrt(b^(2)cos^(2)theta+b^(2)sin^(2)theta))]= a ` or `r^(2)=ab`
Hence, the equation of circle is `x^(2)+y^(2)=ab`