Question

$2x^2+y^2-3xy=0$ be the equation of pair of tangents drawn from the origin O to circles of radius $3$. Find the absolute difference of all possible different lengths of tangents.

• A. $18$
• B. $24$
• C. $30$
• D. None of these

Answer 1

The correct option is A $18$

Given pair of tangent $2x^2+y^2-3xy=0$ from $(0,0)$

Equation of tangents

$2x+(\frac{-3}{2}+\sqrt{\frac{9}{4}-2})y=02x-y=0\left(1\right)2x+(\frac{-3}{2}-\sqrt{\frac{9}{4}-2})y=0x-y=0\left(2\right)$

Let equation of circle $(x-h{)}^2+(y-k{)}^2=3^2$.

Equation of tangent with slope $m$ to circle

$(y-k)=m(x-h)\pm 3\sqrt{1+m^2}y=mx+(k-mh\pm 3)\sqrt{1+m^2}\left(3\right)$

From $\left(1\right),\left(3\right)$

$k-2h\pm 3\sqrt{5}=0\left(4\right)$

From $\left(2\right),\left(3\right)$

$k-h\pm 3\sqrt{2}=0\left(5\right)$

Solving $\left(4\right),\left(5\right)$

$A(3(\sqrt{5}-\sqrt{2}),3(\sqrt{5}-2\sqrt{2})),B(3(\sqrt{2}-\sqrt{5}),3(2\sqrt{2}-\sqrt{5}))C(-3(\sqrt{2}+\sqrt{5}),-3(2\sqrt{2}+\sqrt{5})),D(3(\sqrt{2}+\sqrt{5}),3(2\sqrt{2}+\sqrt{5}))$

Length of tangents $\sqrt{k^2+h^2-9}$

$S_A=\sqrt{3^2{(\sqrt{5}-\sqrt{2})}^2+3^2{(\sqrt{5}-2\sqrt{2})}^2-9}=0.48S_D=\sqrt{3^2{(\sqrt{2}+\sqrt{5})}^2+3^2{(2\sqrt{2}+\sqrt{5})}^2-9}=18.48S_D-S_A=18$