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  • Let the tangents drawn from the origin to the circle,  touch it at the points A and B. The <img al

Let the tangents drawn from the origin to the circle, x^{2}+y^{2}-8x-4y+16=0 touch it at the points A and B. The (AB)^{2} is equal to:
Option: 1  \frac{32}{5}


Option: 2  \frac{64}{5}


Option: 3  \frac{32}{5}


Option: 4  \frac{56}{5}
 

Answers (1)

best_answer

 

 

Tangent from a Point to the Circle (NEW) -

Length of tangent (PT)  from a point to a circle
\begin{array}{l}{\text { The length of the tangent from a point P }\left(x_{1}, y_{1}\right) \text { to the }} \\ {\text { circle } x^{2}+y^{2}=a^{2} \text { is } \sqrt{x_{1}^{2}+y_{1}^{2}-a^{2}}}\end{array}


\\\mathrm{In\;\;\Delta PTC,\;\;PT^2=PC^2-CT^2}\\\text{Here for the circle }x^2+y^2=a^2\;\;\text{coordinate of C is (0, 0)}\\\mathrm{Hence,\;\;PT^2=\left (\sqrt{\left ( x_1-0 \right )^2+\left ( y_1-0 \right )^2} \right )^2-\left (\sqrt{a^2} \right )^2}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;PT=\sqrt{x_1^2+y_1^2-a^2}\;\;\;\;\;\;\;\;\;\;\;[\because CT=radius\;=a]}
\begin{array}{l}{\text { The length of the tangent from a point P }\left(x_{1}, y_{1}\right) \text { to the circle }} \\ {x^{2}+y^{2}+2 g x+2 f y+c=0 \text { is } \sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}}\end{array}

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Length of the tangent from the origin (0,0) is L=\sqrt{S_{1}}=\sqrt{16}=4

The radius of the circleR=\sqrt{16+4-16}=2

Length of Chord of contact \frac{2 L R}{\sqrt{L^{2}+R^{2}}}=\frac{2 \times 4 \times 2}{\sqrt{16+4}}=\frac{16}{\sqrt{20}}

Correct Option (2)

Posted by

Ritika Jonwal

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