The length of radii of two circles are 8 cm and 3 cm and distance between two centres is 13 cm. Find the length of a firect common tangent of two circles.

Explanation

Let the radii of the circles with centres A and B are 8 cm and 3 cm respectively and the distance PQ is the direct common tangent of both the ciecles.
We have to find the length of PQ. Let us join A, P and B, Q and let us draw BM perpendicular to AP from B. Then PQBM is a reactanglt. `[because angle APQ =angle BQP =90^(@)]`
`therefore PQ = BM and PM = BQ.`
As per question, `AP = 8cm, BQ=3cm.` Then
`AM =AP -PM =8cm -BQ=8cm -3 cm =5cm`
Now, in the right-angled tringle ABM we get by pythagoras theorem,
`BM^(2)+AM^(2)=AB^(2)[becauseAB=` hypotenuse]
or, `BM^(2)+5^(2)=13^(2)or, BM^(2)=13^(2)-5^(2)=169-25=144`
`thereforeBM =sqrt144=12`
`therefore PQ = 12 cm [because BM=PQ]`
`therefore` the required length of hte direct common tangent of two circles is 12 cm.