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BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Mathematics
Chapter: Circle

BITSAT 2013 · Mathematics · Circle

A pair of tangents are drawn from the origin to the circle x^ 2 +y^...

A pair of tangents are drawn from the origin to the circle

x^{2}+y^{2}+20(x+y)+20=0,

then the equation of the pair of tangent are

Options

A $x^{2}+y^{2}-5 x y=0$
B $x^{2}+y^{2}+2 x+y=0$
C $x^{2}+y^{2}-x y+7=0$
D $2 x^{2}+2 y^{2}+5 x y=0$

Correct answer

D. $2 x^{2}+2 y^{2}+5 x y=0$

Step-by-step solution

Equation of pair of tangents is given by

\mathrm{SS}_{1}=\mathrm{T}^{2}

\begin{array}{l}\text{ or } \quad \mathrm{S}=\mathrm{x}^{2}+\mathrm{y}^{2}+20(\mathrm{x}+\mathrm{y})+20, \mathrm{ S}_{1}=20 \\ \mathrm{ T}=10(\mathrm{x}+\mathrm{y})+20=0 \\ \therefore \quad \mathrm{SS}_{1}=\mathrm{T}^{2}\\ \Rightarrow \quad 20\left(\mathrm{x}^{2}+\mathrm{y}^{2}+20(\mathrm{x}+\mathrm{y})+20\right)=10^{2} \\ \quad(\mathrm{x}+\mathrm{y}+2)^{2} \\ \Rightarrow \quad 4 \mathrm{x}^{2}+4 \mathrm{y}^{2}+10 \mathrm{xy}=0 \Rightarrow 2 \mathrm{x}^{2}+2 \mathrm{y}^{2}+5 \mathrm{xy}=0 \end{array}

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