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  • Tangent drawn from the point P(4,4) to the circle <img alt="\mathrm{x^2+y^2-2 x-2 y-7=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-2%20x-2%20y-7%3D

Tangent drawn from the point P(4,4) to the circle \mathrm{x^2+y^2-2 x-2 y-7=0} meet the circle at A and B. The length of the chord AB equals

Option: 1

4 \sqrt{3}


Option: 2

2 \sqrt{3}


Option: 3

2 \sqrt{6}


Option: 4

3 \sqrt{2}


Answers (1)

best_answer

Given x^2+y^2-2 x-2 y-7=0

\Rightarrow(x-1)^2+(y-1)^2=3^2

\therefore \, \, Centre \, (1,1),\, \, radius =3=O A

Now, P A=P B=\sqrt{16+16-8-8-7}=\sqrt{9}=3

Now, in \triangle A T P, we have

\begin{aligned} & \frac{A T}{A P}=\sin \alpha \\\\ & \Rightarrow A T=A P \sin \alpha \\\\ & \Rightarrow \quad A T=3 \sin 45^{\circ} \\\\ & {\left[\because O A=A P \Rightarrow \alpha=45^{\circ}\right]} \\\\ & \Rightarrow A T=\frac{3}{\sqrt{2}} \\\\ & \therefore A B=2 A T=2 \times \frac{3}{\sqrt{2}}=3 \sqrt{2} \end{aligned}

Posted by

Gautam harsolia

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