Line l touches a circle with centre O at point P. If radius of the circle is 9 cm,

Explanation:- 

Consider a circle with radius 9 cm.

(1) Given that line l is tangent to the circle at point P.

Since P is the point of contact between line l and the circle, OP is perpendicular to line l. Therefore, OP is the radius of the circle and its length is 9 cm.

Also, d(O,P) represents the distance between point O (the center of the circle) and point P. Since P lies on the circle, the distance between O and P is equal to the radius of the circle, which is 9 cm.

(2) Given that d(O,Q) = 8 cm < radius of the circle.

Since the distance between O and Q is less than the radius of the circle, point Q lies inside the circle.

(3) Given that if d(OQ) = 15 cm, then there are two locations of point Q on the line l, one on the left of point P and one on the right of point P.

Since line l is tangent to the circle at point P, the radius OP is perpendicular to line l. Therefore, angle OPQ is a right angle (90°).

In right triangle OPQ, using the Pythagorean theorem, we have:

OQ^2 = OP^2 + PQ^2

where PQ represents the distance between points P and Q.

Solving for PQ, we get:

PQ = sqrt(OQ^2 – OP^2)

Substituting OQ = 15 cm and OP = 9 cm, we get:

PQ = sqrt(15^2 – 9^2) = sqrt(144) = 12 cm

Therefore, the two locations of point Q on line l are at a distance of 12 cm from point P.