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  • The tangents drawn from the origin to the circle <img alt="\mathrm{x^{2}+y^{2}-2 r x-2 h y+h^{2}=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E%7B2%7D&plus;y%5E%7B2%7

The tangents drawn from the origin to the circle \mathrm{x^{2}+y^{2}-2 r x-2 h y+h^{2}=0} are perpendicular if

Option: 1

\mathrm{2 h=r}


Option: 2

\mathrm{h=-r}


Option: 3

\mathrm{r^{2}+h^{2}=1}


Option: 4

\mathrm{2 r^{2}=h^{2}}


Answers (1)

best_answer

The combined equation of the tangents drawn from (0,0) to \mathrm{x^{2}+y^{2}-2 r x-2 hy +h^{2}=0} is

\mathrm{\left(x^{2}+y^{2}-2 r x-2 h y+h^{2}\right) h^{2}=\left(-r x-h y+h^{2}\right)^{2}}

This equation represents a pair of perpendicular straight lines
If coefficient of \mathrm{x^{2}+} coefficient of  \mathrm{y^{2}=0} i.e. \mathrm{2 h^{2}-r^{2}-h^{2}=0 \Rightarrow r^{2}=h^{2}\, or \, r= \pm h}

Hence (B) is the correct answer.

Posted by

jitender.kumar

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