The angle between the tangents to the circle with centre (4, 5) drawn from P(–2, –3) is ${120}^0,$ then length of the tangent to the circle from P is 
 

$$\begin{gathered} \mathrm{Given} \mathrm{centre} \mathrm{O}=(4,5),\mathrm{P}(-2,-3) \\ \mathrm{and} \angle \mathrm{BPA}={120}^0 \\ \Rightarrow \angle \mathrm{APO}=\frac{1}{2}\angle \mathrm{APB}=\frac{1}{2}\left({120}^0\right)={60}^0 \\ \mathrm{OP}=\sqrt{{(4+2)}^2+{(5+3)}^2 } \\ \Rightarrow \mathrm{OP}=10 \\ \mathrm{In} \mathrm{triangle} \mathrm{APO} \cos \angle \mathrm{APO}=\frac{\mathrm{AP}}{\mathrm{PO}} \\ \Rightarrow \cos {60}^0=\frac{\mathrm{AP}}{10} \\ \frac{1}{2}=\frac{\mathrm{AP}}{10} \Rightarrow \mathrm{AP}=5 \\ \therefore \mathrm{length} \mathrm{of} \mathrm{tangent} \mathrm{is} 5 \end{gathered}$$