Circles - Prove lengths of tangents from an external point to a circle are equal

A point $Q$ is at a distance of $25$ cm from the center of a circle and the length of the tangent $QT$ to the circle is $24$ cm. Find the radius of the circle.

1. Let $O$ be the center of the circle and $r$ be the radius.
2. In the right-angled triangle $\triangle OTQ$, $\angle OTQ = 90^{\circ}$ because the radius is perpendicular to the tangent at the point of contact.
3. Using the Pythagoras Theorem: $OQ^2 = OT^2 + QT^2$
4. Substitute the given values: $25^2 = r^2 + 24^2$
5. $625 = r^2 + 576$
6. $r^2 = 625 - 576 = 49$
7. $r = \sqrt{49} = 7$ cm.

We identify the right-angled triangle formed by the radius, the tangent, and the line from the center to the external point, then apply the Pythagoras theorem to find the missing side.

Two tangents $PA$ and $PB$ are drawn to a circle with center $O$ from an external point $P$. If $\angle AOB = 110^{\circ}$, then find the measure of $\angle APB$.

1. In quadrilateral $OAPB$, $\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$ because the radius is perpendicular to the tangent at the point of contact.
2. The sum of interior angles of a quadrilateral is $360^{\circ}$.
3. Therefore, $\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^{\circ}$.
4. Substituting the values: $90^{\circ} + \angle APB + 90^{\circ} + 110^{\circ} = 360^{\circ}$.
5. $290^{\circ} + \angle APB = 360^{\circ}$.
6. $\angle APB = 360^{\circ} - 290^{\circ} = 70^{\circ}$.

This problem uses the property that the angles formed by the radii and tangents at the points of contact are $90^{\circ}$, and that the angle at the center and the angle between tangents are supplementary.

Let $\angle PTQ = \theta$. In $\triangle TPQ$, we have $TP = TQ$ (Tangents from an external point). So, $\triangle TPQ$ is an isosceles triangle. Therefore, $\angle TPQ = \angle TQP = \frac{1}{2}(180^{\circ} - \theta) = 90^{\circ} - \frac{\theta}{2}$. We know that $OP \perp TP$ (Radius is perpendicular to tangent). So, $\angle OPT = 90^{\circ}$. Now, $\angle OPQ = \angle OPT - \angle TPQ$ $\angle OPQ = 90^{\circ} - (90^{\circ} - \frac{\theta}{2}) = \frac{\theta}{2}$. Hence, $\angle OPQ = \frac{1}{2}\angle PTQ$, which implies $\angle PTQ = 2\angle OPQ$.

This proof utilizes the property that tangents from an external point are equal in length, forming an isosceles triangle, and the radius is perpendicular to the tangent at the point of contact.

Let $TR = y$ and $TP = x$. Let $O$ be the center. $OR \perp PQ$ and $R$ bisects $PQ$. So, $PR = RQ = 4$ cm. In $\triangle OPR$, $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3$ cm. In $\triangle PRT$, $x^2 = y^2 + 4^2$ (Pythagoras theorem) ... (1) In $\triangle OPT$, $OT^2 = TP^2 + OP^2$ (Tangent $\perp$ Radius) $(y + 3)^2 = x^2 + 5^2$ ... (2) Subtracting (1) from (2): $(y + 3)^2 - y^2 = 5^2 - 4^2$ $y^2 + 6y + 9 - y^2 = 25 - 16$ $6y + 9 = 9 \implies 6y = 16 - 9$? No, calculation: $y^2+6y+9-y^2 = 25-16 = 9$, so $6y=0$? Re-check. Correct approach: $\triangle TRP \sim \triangle PRO$ is incorrect. $\triangle TRP \sim \triangle TPO$ is correct. $\frac{TP}{PO} = \frac{RP}{RO} \implies \frac{x}{5} = \frac{4}{3} \implies x = \frac{20}{3}$ cm.

By using similarity between $\triangle TRP$ and $\triangle PRO$ (or using Pythagoras in two related triangles), the length of the tangent can be calculated.