The radii of two circles are R and r unit `(R gtr).` If the distance between the two centres of the circles be d unit, then prove that the length of their transversal common tangent `=sqrt(d^(2) -(R+r)^(2))` unit.

Explanation

Let the radii of the circle with centres A and B be R unit and r unit respectively PQ is a transversal common tangent to the circle which inersect the two circles at the points P and Q respectively. A perpendicular BM is drawn from B to the extended AP.
To prove : `PQ = sqrt(d^(2) -(R+r)^(2))` where d= distance between the centres of the circel = AB.
Proof : `because angle BQP =angle QPM =90^(@)and BQ -PM`
`therefore PQ =BM and DeltaAMB` is a right-angled triangle of which the hypotenuse is AB.
`therefore` from Pythagoras theorem we get,
`AB^(2)=AM^(2)+BM^(2)`
`or, d^(2) =(AP +PM)^(2) +PQ^(2)[because AB=d , BM=PQ]`
`or , d^(2) =(AP +BQ)^(2)+PQ^(2)[becausePM=BQ]`
`or, d^(2)=(R+r)^(2)+PQ^(2)[because AP=R, BQ=r]`
`or, PQ^(2)=d^(2) -(R+r)^(2)`
`or, PQ =sqrt(d^(2) -(R+r)^(2))`
Hence `PQ=sqrt(d^(2)-(R+r)^(2))` unit. (Proved)]