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Let A B C be a triangle right-angled at ...

Let `A B C` be a triangle right-angled at `Aa n dS` be its circumcircle. Let `S_1` be the circle touching the lines `A B` and `A C` and the circle `S` internally. Further, let `S_2` be the circle touching the lines `A B` and `A C` produced and the circle `S` externally. If `r_1` and `r_2` are the radii of the circles `S_1` and `S_2` , respectively, show that `r_1r_2=4` area `( A B C)dot`

A

`(r_(1)r_(2))/4`

B

`(r_(1)r_(2))/8`

C

`(r_(1)r_(2))/2`

D

`r_(1)r_(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Equation of `S=(x-a/2)^(2)+(y-b/2)^(2)=(a^(2)+b^(2))/4`
`x^(2)+y^(2)-ax-by=0`
And circle with centre `(r,r)` and radius `r`
Touches `AB` and `AC` its equation is
`(x-r)^(2)+(y-r)^(2)=r^(2)`
`x^(2)+y^(2)-2xr-2yr+r^(2)=0`
Both circle touches internally or externally
Using `sqrt((r-a/2)^(2)+(r-b/2)^(2))=sqrt(((a^(2))/4+(b^(2))/4))+-r`
`DeltaABC=(r_(1)r_(2))/4`
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