A line meets the co-ordinate axes in $A$ and $B$ . A circle is circumscribed about the triangle $O A B$ , $d_{1}$ and $d_{2}$ are the distance of the tangent to the circle at the origin $O$ from the point $A$ and respectively and diameter of the circle is $λ_{1} d_{1} + λ_{2} d_{2}$ , then find the value of $λ_{1} + λ_{2}$ .

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Solution

$E D = d_{2}$
$F D = d_{1}$
$d n △ A F D$ & $△ C O D$
$\frac{x}{d_{1}} = \frac{x + r}{r} \Rightarrow \frac{x}{x + r} = \frac{d_{1}}{r} - - - - - - - - (1)$
In $△ A F D$ & $△ B E D$
$\frac{x}{d_{1}} = \frac{x + 2 r}{d_{2}}$
$x = \frac{d_{1}}{d_{2}} (x + 2 r) \Rightarrow x (1 - \frac{d_{1}}{d_{2}}) \Rightarrow \frac{2 r d_{1}}{d_{2}}$
$x = \frac{2 r d_{1}}{d_{2} - d_{1}} \Rightarrow \frac{x}{x + r} = \frac{d_{1}}{r} \Rightarrow \frac{2 r d_{1}}{(d_{2} - d_{1}) (\frac{2 r d_{1}}{d_{2} - d_{1}} + r)} = \frac{d_{2}}{r}$
$2 r^{2} = d_{2} - d_{1} [\frac{2 r d_{1} + r d_{2} - r d_{1}}{d_{2} - d_{1}}]$
$2 r^{2} = r [d t d_{2}]$
$r = \frac{d_{1} + d_{2}}{2}$

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A line meets the co-ordinate axes in A and B. A circle is circumscribed about the triangle OAB, d1 and d2 are the distance of the tangent to the circle at the origin O from the point A and respectively and diameter of the circle is λ1d1+λ2d2, then find the value of λ1+λ2.

ED=d2FD=d1dn△AFD & △CODxd1=x+rr⇒xx+r=d1r−−−−−−−−(1)In △AFD & △BEDxd1=x+2rd2x=d1d2(x+2r)⇒x(1−d1d2)⇒2rd1d2x=2rd1d2−d1⇒xx+r=d1r⇒2rd1(d2−d1)(2rd1d2−d1+r)=d2r2r2=d2−d1[2rd1+rd2−rd1d2−d1]2r2=r[dtd2]r=d1+d22