A secant is drawn from a point P to a circle so that it meets the circle first at A, then goes through the centre, and leaves the circle at B. If the length of the tangent from P to the circle is 12 cm, and the radius of the circle is 5 cm, then the distance from P to A is:

Given,

Radius of the circle = 5 cm

Let the distance from P to A = a

PB is the secant and PV is the tangent to the circle from external point P

$$=$$>  $$PV^2=PA.PB$$

$$=$$>  $$12^2=PA\left(PA+AB\right)$$

$$=$$>  $$144=a\left(a+10\right)$$

$$=$$>  $$144=a^2+10a$$

$$=$$>  $$a^2+10a-144=0$$

$$=$$>  $$a^2+18a-8a-144=0$$

$$=$$>  $$a\left(a+18\right)-8\left(a+18\right)=0$$

$$=$$>  $$\left(a+18\right)\left(a-8\right)=0$$

$$=$$>  $$a+18=0$$  or   $$a-8=0$$

$$=$$>  $$a=-18$$  or   $$a=8$$

a cannot be negative

$$=$$>  $$a=8$$ cm

$$\therefore\ $$Distance from P to A = 8 cm

Hence, the correct answer is Option B