If the length of a common internal tangent to two circles is 7 , and that of a common external tangent is 11 , then the product of the radii of the two circles is:

Question

If the length of a common internal tangent to two circles is 7 , and that of a common external tangent is 11 , then the product of the radii of the two circles is: (A) 36 (B) 9 (C) 18 (D) 4

Tardigrade Admin · Accepted Answer

as we know L text int =√d2-(r1+r2)2=7 Le x t=√d2-(r1-r2)2=11 squaring subtact r1 r2=18

Q. If the length of a common internal tangent to two circles is 7 , and that of a common external tangent is 11 , then the product of the radii of the two circles is:

36

9

18

4

as we know $L_{int} = d^{2} - (r_{1} + r_{2})^{2} = 7$
$L_{e x t} = d^{2} - (r_{1} - r_{2})^{2} = 11$
squaring & subtact $r_{1} r_{2} = 18$

Q. If the length of a common internal tangent to two circles is 7 , and that of a common external tangent is 11 , then the product of the radii of the two circles is:

36

9

18

4

as we know Lint ​=d2−(r1​+r2​)2​=7 Lext​=d2−(r1​−r2​)2​=11 squaring & subtact r1​r2​=18

as we know $L_{int} = d^{2} - (r_{1} + r_{2})^{2} = 7$
$L_{e x t} = d^{2} - (r_{1} - r_{2})^{2} = 11$
squaring & subtact $r_{1} r_{2} = 18$