If `a >2b >0,`
then find the positive value of `m`
for which `y=m x-bsqrt(1+m^2)`
is a common tangent to `x^2+y^2=b^2`
and `(x-a)^2+y^2=b^2dot`
Text Solution
Verified by Experts
`y=mx-b sqrt(1+m^(2))` is a tangnet to the circle `x^(2)+y^(2)=b^(2)` for all vales of m. If it also touches the circle `(x-a)^(2)+y^(2)=b^(2)` , then the length of the perpendicular from tis center (a,0) on this line is equal to the radiu b of the cirle, which give `(ma-bsqrt(1+m^(2)))/(sqrt(1+m^(2)))=+-b` Taking the negative value of the RHS, we get `m=0` . So, we neglect it. Taking the positive value of the RHS, we get `ma=2bsqrt(1+m^(2))` or `m^(2)(a^(2)-4b^(2))=4b^(2)` or `m=(2b)/(sqrt(a^(2)-4b^(2)))`
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