Circles

A circle can have infinitely many tangents, because a tangent can be drawn at every point on the circle, and there is one and only one tangent at each point.

From the chapter:
- A tangent intersects a circle in one point.
- A line intersecting a circle in two points is called a secant.
- A circle can have two parallel tangents at the most.
- The common point of a tangent and the circle is called the point of contact.

Join $OP$. Since $PQ$ is tangent at $P$, $OP \perp PQ$.
So in right triangle $OPQ$:
$$OP=5\text{ cm},\quad OQ=12\text{ cm}$$
By Pythagoras theorem,
$$PQ^2=OQ^2-OP^2=12^2-5^2=144-25=119$$
$$PQ=\sqrt{119}\text{ cm}$$
The printed option matching this is (D).

Let the radius be $r$. The tangent length from the external point is 24 cm and the distance from the centre is 25 cm.
Using the right triangle formed by radius to the point of contact:
$$r^2+24^2=25^2$$
$$r^2=625-576=49$$
$$r=7\text{ cm}$$
So the correct option is (A).

For two tangents from an external point,
$$\angle PTQ + \angle POQ = 180^\circ$$
Given $\angle POQ=110^\circ$,
$$\angle PTQ = 180^\circ-110^\circ=70^\circ$$
So the correct option is (B).

Let the angle between the tangents be
$$\angle APB=80^\circ$$
From the chapter, the centre lies on the bisector of the angle between the tangents, so $OP$ bisects $\angle APB$.
Hence
$$\angle APO = 40^\circ$$
Now in right triangle $OAP$, since radius is perpendicular to tangent,
$$\angle OAP=90^\circ$$
Therefore,
$$\angle POA = 180^\circ - 90^\circ - 40^\circ = 50^\circ$$
So the correct option is (A).

Let the circle have centre $O$ and let the diameter be $AB$. Tangents are drawn at $A$ and $B$.

By Theorem 10.1, the tangent at $A$ is perpendicular to radius $OA$, and the tangent at $B$ is perpendicular to radius $OB$.

Since $A, O, B$ are collinear on a diameter,
$$OA \text{ and } OB$$
lie on the same straight line.

So both tangents are perpendicular to the same line $AB$.

Lines perpendicular to the same line are parallel.

Hence, the tangents drawn at the ends of a diameter are parallel.

Let the tangent touch the circle at point $P$ and let $O$ be the centre.

By Theorem 10.1, the tangent at any point of a circle is perpendicular to the radius through the point of contact. So if a line is drawn perpendicular to the tangent at $P$, then it is along the radius $OP$.

Since the radius passes through the centre, the perpendicular at the point of contact to the tangent passes through the centre.