Circles - Introduction

From a point $P$, the length of the tangent to a circle is $24$ cm and the distance of $P$ from the center is $25$ cm. Find the radius of the circle.

Let $O$ be the center and $T$ be the point of contact. Given $PT = 24$ cm and $OP = 25$ cm. Since radius $OT \perp PT$, $\triangle OTP$ is a right-angled triangle. Using Pythagoras Theorem: $OT^2 + PT^2 = OP^2$ $OT^2 + 24^2 = 25^2$ $OT^2 + 576 = 625$ $\begin{array}{r} 625 \\ -576 \\ \hline 49 \end{array}$ $OT^2 = 49$ $OT = \sqrt{49} = 7 \text{ cm}$

The radius is calculated using the property that the radius is perpendicular to the tangent at the point of contact, forming a right triangle with the distance from the center.

If tangents $PA$ and $PB$ from a point $P$ to a circle with center $O$ are inclined to each other at an angle of $80^\circ$, then find $\angle POA$.

Given $\angle APB = 80^\circ$. Since $PA$ and $PB$ are tangents, $OA \perp PA$ and $OB \perp PB$. In quadrilateral $OAPB$, $\angle AOB + \angle APB = 180^\circ$. $\angle AOB + 80^\circ = 180^\circ$ $\angle AOB = 100^\circ$ In $\triangle OAP$ and $\triangle OBP$, $OP$ is common and $PA=PB$, so $\triangle OAP \cong \triangle OBP$. Thus, $OP$ bisects $\angle AOB$: $\angle POA = \frac{1}{2} \angle AOB$ $\angle POA = \frac{100^\circ}{2} = 50^\circ$

First, we find the central angle $\angle AOB$ using the supplementary property of tangents. Then, we use the symmetry of the tangents from an external point to find that $\angle POA$ is half of the central angle.

Let $OT$ intersect $PQ$ at $R$. Since $OT$ is the angle bisector of $\angle PTQ$, $OT \perp PQ$. Thus, $PR = RQ = 4$ cm. In $\triangle ORP$, by Pythagoras Theorem: $OR^2 = OP^2 - PR^2 = 5^2 - 4^2 = 25 - 16 = 9$ $OR = 3 \text{ cm}$ Now, in right $\triangle TRP$ and $\triangle PRO$: $\angle TPR + \angle RPO = 90^\circ$ $\angle POR + \angle RPO = 90^\circ$ So, $\angle TPR = \angle POR$ Thus, $\triangle TRP \sim \triangle PRO \text{ (By AA similarity)}$ $\frac{TP}{PO} = \frac{RP}{RO}$ $\frac{TP}{5} = \frac{4}{3}$ $TP = \frac{20}{3} \approx 6.67 \text{ cm}$

We use the property that the line joining the center to the external point is the perpendicular bisector of the chord joining the points of contact. We then apply Pythagoras theorem to find OR and use similar triangles (TRP and PRO) to find the tangent length TP.

Let the points of contact be $P, Q, R, \text{ and } S$ on sides $AB, BC, CD, \text{ and } DA$ respectively. Using the property that lengths of tangents from an external point are equal:

1. $AP = AS$
2. $BP = BQ$
3. $CR = CQ$
4. $DR = DS$

Adding these equations: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$ $AB + CD = AD + BC$ Hence proved.

This result stems from the equality of tangent segments from each vertex of the quadrilateral to the circle. By grouping segments that form the opposite sides, we see the sums are identical.