Tangent drawn from a point P, touches the circle at Q. O is the centre of this circle. If PQ = 12 cm and OQ = 5 cm, then what is the value of OP?

15 cm $\sqrt{119} $ cm $\sqrt{117 } $ cm 13 cm

13 cm

According tot the concept, \(\angle\)OQP = \({90}^\circ\) Hence, OP is the hypotenuse of \(\Delta \)OQP which is a right angled triangle. Now, OP = \(\sqrt { {12 }^{2 }  + {5 }^{2 }  }\) = 13 cm Therefore, OP is 13 cm.