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The equation of the tangents drawn from the origin to the circle x^2+y^2-2 r x-2 h y+h^2=0 ,are

Option: 1

x=0


Option: 2

y=0


Option: 3

\left(h^2-r^2\right) x-2 r h y=0


Option: 4

\left(h^2-r^2\right) x+2 r h y=0


Answers (1)

best_answer

The given equation is 

                  (x-r)^2+(y-h)^2=r^2

Tangents are x=0

                   

and               

             \begin{aligned} y= & x \tan \left(\frac{\pi}{2}-2 \alpha\right) \\ \\= & x \cot 2 \alpha \\ \\& =\frac{x\left(1-\tan ^2 \alpha\right)}{2 \tan \alpha} \\ \\& y=\frac{x\left(1-\frac{r^2}{h^2}\right)}{2\left(\frac{r}{h}\right)} \quad\left(\because \text { in } \triangle O D C, \tan \alpha=\frac{r}{h}\right) \end{aligned}   

Or                \left(h^2-r^2\right) x-2 r h y=0

Posted by

Devendra Khairwa

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