Circles - Apply tangent properties to solve geometry and mensuration-based problems

A point $P$ is $13\text{ cm}$ away from the center of a circle. If the length of the tangent drawn from $P$ to the circle is $12\text{ cm}$, find the radius of the circle.

1. Let $O$ be the center of the circle and $T$ be the point of contact.
2. In $\triangle OTP$, the radius $OT$ is perpendicular to the tangent $PT$, so $\angle OTP = 90^{\circ}$.
3. Using the Pythagorean Theorem: $OP^2 = OT^2 + PT^2$.
4. Substitute the given values: $13^2 = r^2 + 12^2$.
5. $169 = r^2 + 144$.
6. $r^2 = 169 - 144 = 25$.
7. $r = \sqrt{25} = 5\text{ cm}$.

This problem uses the property that the radius is perpendicular to the tangent at the point of contact, creating a right-angled triangle where the distance from the center is the hypotenuse.

Two tangents $PA$ and $PB$ are drawn to a circle with center $O$ from an external point $P$. If $\angle APB = 80^{\circ}$, calculate the value of $\angle POA$.

1. We know that the line joining the external point $P$ to the center $O$ bisects the angle between the tangents.
2. Therefore, $\angle APO = \frac{1}{2} \angle APB = \frac{1}{2} \times 80^{\circ} = 40^{\circ}$.
3. In $\triangle OAP$, we know $\angle OAP = 90^{\circ}$ (radius perpendicular to tangent).
4. The sum of angles in $\triangle OAP$ is $180^{\circ}$.
5. $\angle POA + \angle OAP + \angle APO = 180^{\circ}$.
6. $\angle POA + 90^{\circ} + 40^{\circ} = 180^{\circ}$.
7. $\angle POA = 180^{\circ} - 130^{\circ} = 50^{\circ}$.

This solution relies on two properties: first, that the line from the center to the external point bisects the angle between the tangents; and second, the angle sum property of the right-angled triangle formed by the radius and tangent.

Let the circle touch the sides $AB$, $BC$, $CD$, and $DA$ at points $P$, $Q$, $R$, and $S$ respectively. Since tangents from an external point are equal: $AP = AS$ (Tangents from $A$) $BP = BQ$ (Tangents from $B$) $CR = CQ$ (Tangents from $C$) $DR = DS$ (Tangents from $D$) Adding these equations: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$ $AB + CD = AD + BC$.

This problem uses the property that lengths of tangents from an external point to a circle are equal. By identifying the four sets of equal tangents and grouping them according to the sides of the quadrilateral, we arrive at the proof.

Join $OC$. In $\triangle OPA$ and $\triangle OCA$: $OP = OC$ (Radii) $AP = AC$ (Tangents from $A$) $OA = OA$ (Common) $\triangle OPA \cong \triangle OCA \Rightarrow \angle POA = \angle COA$ ... (i) Similarly, $\triangle OQB \cong \triangle OCB \Rightarrow \angle QOB = \angle COB$ ... (ii) Since $POQ$ is a diameter, it is a straight line: $\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$ $2\angle COA + 2\angle COB = 180^{\circ}$ $\angle COA + \angle COB = 90^{\circ}$ $\angle AOB = 90^{\circ}$.

By proving the congruence of triangles formed by the radii and tangents, we establish that the central angle is bisected. Since the total angle along the diameter is 180 degrees, the sum of the bisected parts must be 90 degrees.