10.2-Number of Tangents from a Point on a Circle

10.2-Number of Tangents from a Point on a Circle Important Formulae

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Grade 10 → Math → Circles → 10.2-Number of Tangents from a Point on a Circle

After successful completion of this topic, you should be able to:

  • Prove and apply theorems related to tangent of a circle in order to determine number of tangents from the given point(s).
  • Prove and apply theorem related to tangent of a circle in order to determine length of the tangent.

In this section, we will explore the concept of tangents to a circle from a given point. A tangent is a straight line that touches the circle at exactly one point. Understanding the number of tangents that can be drawn from a point outside a circle is fundamental in geometry.

Tangents from an External Point

Let us consider a circle with center $O$ and radius $r$. If a point $P$ lies outside the circle, we can determine the number of tangents that can be drawn from point $P$ to the circle. The following scenarios arise:

  • If point $P$ is outside the circle, there are exactly two tangents that can be drawn from point $P$ to the circle.
  • If point $P$ is on the circle, there is one tangent that can be drawn at that point.
  • If point $P$ is inside the circle, no tangents can be drawn from point $P$ to the circle.

In the diagram:

  • Circle with center $O$ and radius $r$.
  • Point $P$ outside the circle, where two tangents $PA$ and $PB$ touch the circle at points $A$ and $B$, respectively.

Distance from the Center to the Point

The distance from the center $O$ of the circle to the external point $P$ is crucial in determining the number of tangents. Let the distance be $d$. The relationship between $d$, the radius $r$, and the tangents can be summarized as follows:

  • If $d > r$, then there are two tangents.
  • If $d = r$, then there is one tangent.
  • If $d < r$, then there are no tangents.

Tangent Length Formula

The length of the tangent from an external point $P$ to the circle can also be calculated using the following formula:

$L = \sqrt{d^2 - r^2}$

Where:

  • $L$ is the length of the tangent.
  • $d$ is the distance from the point to the center of the circle.
  • $r$ is the radius of the circle.

Example Problem

Consider a circle with radius $r = 5$ units and an external point $P$ located $d = 8$ units from the center $O$. To find the number of tangents and their lengths:

  1. Since $d > r$, there are two tangents from point $P$.
  2. Using the tangent length formula:

    3. $L = \sqrt{d^2 - r^2} = \sqrt{8^2 - 5^2} = \sqrt{64 - 25} = \sqrt{39}$ units

Thus, the lengths of both tangents from point $P$ to the circle are $\sqrt{39}$ units each.

Summary of Key Points

  • Two tangents can be drawn from a point outside the circle.
  • One tangent can be drawn if the point lies on the circle.
  • No tangents can be drawn from a point inside the circle.

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:

(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm

Solution:

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:


(A) 7 cm
(B) 12 cm
(C) 15 cm CORRECT
(D) 24.5 cm

Solution:

Let the radius of the circle be $r$ cm. We are given: - The length of the tangent $QT = 24$ cm - The distance from Q to the center $OQ = 25$ cm We know that the radius of the circle is perpendicular to the tangent at the point of contact. Therefore, triangle $OQT$ is a right triangle, with $OQ$ as the hypotenuse, $OT$ as the radius, and $QT$ as the tangent. By the Pythagorean theorem:
$$OQ^2 = OT^2 + QT^2$$
$$25^2 = r^2 + 24^2$$
$$625 = r^2 + 576$$
$$r^2 = 625 - 576$$
$$r^2 = 49$$
$$r = \sqrt{49} = 7$$
Therefore, the radius of the circle is 7 cm.

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In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that POQ = 110°, then PTQ is equal to:
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:

Question: In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that POQ = 110°, then PTQ is equal to:
(A) 60°
(B) 70°
(C) 80°
(D) 90°
CORRECT: (B) 70°

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If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to:

(A) 50°
(B) 60°
(C) 70°
(D) 80°

Solution:

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to:


(A) 50° CORRECT
(B) 60°
(C) 70°
(D) 80°

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Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let the circle be centered at the origin O with radius r. Consider a circle with center O and radius r. Let A and B be the endpoints of a diameter of the circle. 1. Equation of the circle: The equation of the circle with center O and radius r is given by: $x^2 + y^2 = r^2$ 2. Tangents at points A and B: The tangents at points A and B are perpendicular to the radius of the circle at those points. Let the coordinates of point A be $(r, 0)$ and the coordinates of point B be $(-r, 0)$. The tangent at point A is perpendicular to the radius OA, and the tangent at point B is perpendicular to the radius OB. The slopes of these tangents can be found as follows: 3. Slope of the tangent at A: The radius OA is a horizontal line along the x-axis. The tangent at A is perpendicular to the radius, so it will be a vertical line. The slope of a vertical line is undefined. 4. Slope of the tangent at B: Similarly, the radius OB is a horizontal line along the x-axis, and the tangent at B is also perpendicular to OB, which will be a vertical line. The slope of this vertical line is also undefined. Since both tangents are vertical lines, they are parallel to each other. Thus, we have proved that the tangents drawn at the ends of a diameter of a circle are parallel.

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Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Solution:

Proof: Perpendicular at the point of contact to the tangent passes through the centre
Let the circle be centered at $O$ with radius $r$. Let $P$ be the point of contact of the tangent to the circle, and let $T$ be the point where the tangent touches the circle. 1. Draw the radius $OP$ from the center of the circle to the point of contact $P$. 2. Draw the tangent line at $P$ which touches the circle only at $P$. Let this tangent be $TP$. 3. According to the property of tangents, the radius at the point of contact is perpendicular to the tangent. Thus, $OP \perp TP$. 4. Now, consider the line joining the center $O$ to the point of contact $P$, which is the radius $OP$. 5. Since $OP \perp TP$, the line $OP$ is perpendicular to the tangent $TP$. 6. Hence, the perpendicular drawn from the center of the circle to the tangent at the point of contact passes through the point of contact and the center of the circle. Therefore, the perpendicular at the point of contact to the tangent passes through the center of the circle.

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The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution:

Question:

The length of a tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution:

Let the centre of the circle be O and the point of contact of the tangent be P. The given information is:

  • Distance of point A from the centre O = 5 cm
  • Length of the tangent AP = 4 cm

In the right-angled triangle OAP, where OP is the radius of the circle (let the radius be $r$), and AP is the length of the tangent. According to the Pythagorean theorem:

We have the relation: $OA^2 = OP^2 + AP^2$

Substitute the known values: $5^2 = r^2 + 4^2$

Which simplifies to: $25 = r^2 + 16$

Now, solve for $r^2$: $r^2 = 25 - 16 = 9$

Thus, $r = \sqrt{9} = 3$ cm.

The radius of the circle is 3 cm.

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Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:

Question: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Let the radii of the larger and smaller circles be $R = 5 \, \text{cm}$ and $r = 3 \, \text{cm}$ respectively.

Let the center of the circles be $O$. Let the chord $AB$ of the larger circle touch the smaller circle at point $P$. We are to find the length of the chord $AB$.

Since the chord touches the smaller circle, the distance from the center $O$ to the chord $AB$ is equal to the radius of the smaller circle, i.e., $OP = r = 3 \, \text{cm}$.

Let $M$ be the midpoint of the chord $AB$. Then, $OM$ is perpendicular to the chord $AB$, and $OM = OP = 3 \, \text{cm}$.

In the right-angled triangle $OMA$, using the Pythagorean theorem:

$OA^2 = OM^2 + AM^2$

Since $OA = R = 5 \, \text{cm}$ and $OM = 3 \, \text{cm}$, we have:

$5^2 = 3^2 + AM^2$

$25 = 9 + AM^2$

$AM^2 = 25 - 9 = 16$

$AM = 4 \, \text{cm}$

Since $M$ is the midpoint of the chord $AB$, the length of the chord $AB$ is $2 \times AM = 2 \times 4 = 8 \, \text{cm}$.

The length of the chord $AB$ is $8 \, \text{cm}$.

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A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that: AB + CD = AD + BC .

Solution:

Prove that AB + CD = AD + BC for a quadrilateral circumscribed about a circle

Let the quadrilateral ABCD be drawn to circumscribe a circle. By the property of a quadrilateral that circumscribes a circle, the sum of the lengths of opposite sides is equal. In this case, we have:

Let the points of tangency of the incircle be P, Q, R, and S, where:

  • P is the point of tangency on side AB,
  • Q is the point of tangency on side BC,
  • R is the point of tangency on side CD,
  • S is the point of tangency on side DA.

Let the lengths of the tangents drawn from a point to a circle be equal. Therefore, the lengths of the tangents from each vertex to the points of tangency will be:

  • AP = AS,
  • BP = BQ,
  • CQ = CR,
  • DR = DS.

Now, express the sides of the quadrilateral in terms of these tangents:

  • AB = AP + BP = AS + BQ,
  • BC = BQ + CQ = BP + CR,
  • CD = CR + DR = CQ + DS,
  • DA = DS + AS = DR + AP.

From the above, we can write:

  • AB + CD = (AP + BP) + (CR + DR),
  • AD + BC = (DS + AS) + (BP + CQ).

Using the equal tangents property, we simplify both sides:

  • AB + CD = (AS + BQ) + (CQ + DS),
  • AD + BC = (DS + AS) + (BP + CR).

Thus, we have shown that:

AB + CD = AD + BC.

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In Fig. 10.13, XY and X Y are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X Y at B. Prove that AOB = 90°.

Solution:

Prove that $\angle AOB = 90^\circ$

Given: In Fig. 10.13, XY and XY are two parallel tangents to a circle with center O, and another tangent AB with point of contact C intersects XY at A and XY at B.

To Prove: $\angle AOB = 90^\circ$

Proof:

  1. Let the circle have center O, and XY, XY be the two parallel tangents to the circle.
  2. Let AB be the third tangent to the circle with point of contact C. The tangents from an external point are equal in length. Therefore, we have:
    • OA = OB (radii of the circle)
    • AC = BC (tangents from the point B)
  3. Since XY and XY are parallel, the angles $\angle OAX$ and $\angle OBY$ are right angles (tangent at a point is perpendicular to the radius). Thus:
    • $\angle OAX = 90^\circ$ and $\angle OBY = 90^\circ$
  4. Now, considering triangles OAC and OBC:
    • We have OA = OB (radii of the same circle),
    • AC = BC (equal tangents from the same external point B),
    • and $\angle OAC = \angle OBC = 90^\circ$ (since tangents are perpendicular to the radius at the point of contact).
  5. Hence, $\triangle OAC \cong \triangle OBC$ by RHS (Right angle-Hypotenuse-Side) congruence rule.
  6. Since the triangles are congruent, we conclude that:
    • AO = BO, and thus, $\angle AOB = 90^\circ$

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Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Solution:

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Let the given circle have centre $O$ and radius $r$. Let $P$ be an external point from which two tangents $PA$ and $PB$ are drawn to the circle. The points of contact of the tangents with the circle are $A$ and $B$, respectively. The line-segment joining the points of contact is $AB$. We are required to prove that the angle between the two tangents, $\angle APB$, is supplementary to the angle subtended by the line-segment $AB$ at the centre, $\angle AOB$.

We know the following properties:

  • The radius of the circle is perpendicular to the tangent at the point of contact. Therefore, $OA \perp PA$ and $OB \perp PB$.
  • The tangents from an external point to a circle are equal in length. Therefore, $PA = PB$.
  • In triangle $OAP$ and triangle $OBP$, we have $OA = OB$ (radii of the same circle), $PA = PB$ (tangents from an external point), and $\angle OAP = \angle OBP = 90^\circ$ (since radius is perpendicular to tangent). Hence, $\triangle OAP \cong \triangle OBP$ by RHS congruence.

Therefore, $\angle OAP = \angle OBP$. Also, $\angle APB = 180^\circ - (\angle OAP + \angle OBP)$. Since $\angle OAP = \angle OBP$, we have:

$$\angle APB = 180^\circ - 2 \times \angle OAP$$

Now, consider $\angle AOB$. Since the tangents are equal, $\triangle OAP$ and $\triangle OBP$ are congruent, and hence $\angle AOB = 2 \times \angle OAP$.

Thus, $\angle APB + \angle AOB = 180^\circ$, which proves that the angle between the two tangents is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

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Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

Prove that the parallelogram circumscribing a circle is a rhombus

Let ABCD be a parallelogram that circumscribes a circle with center O. Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

Since the circle is inscribed in the parallelogram, the tangents from any point to the circle are equal in length. Therefore, the following relations hold:

  • AP = AS
  • BP = BQ
  • CQ = CR
  • DR = DS

Now, in a parallelogram, opposite sides are equal. So, we have:

  • AB = CD
  • BC = DA

Let's consider the sum of the lengths of the sides of the parallelogram:

  • AB + BC = CD + DA

Substitute the length of the sides in terms of the tangents:

  • (AP + BP) = (CR + DR)

Using the equal tangents, we get:

  • (AS + BQ) = (CR + DS)

Therefore, all sides of the parallelogram are equal, meaning ABCD is a rhombus.

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A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Solution:

Question 11

In triangle ABC, a circle is drawn to be the incircle with radius 4 cm. The incircle touches side BC at D, dividing it into segments BD = 8 cm and DC = 6 cm. We are asked to find the lengths of sides AB and AC.

Let the lengths of sides AB, BC, and AC be denoted as $c$, $a$, and $b$ respectively. The points where the incircle touches the sides are as follows:

  • BD = s - b = 8 cm
  • DC = s - c = 6 cm

Let the semi-perimeter of the triangle be $s$. From the given information:

  • BD = s - b = 8 cm → $b = s - 8$
  • DC = s - c = 6 cm → $c = s - 6$

We know that the total length of side BC is $a = BD + DC = 8 + 6 = 14$ cm.

Now, the semi-perimeter $s$ is given by the formula:

$$s = \frac{a + b + c}{2}$$

Substituting the values of $b$ and $c$:

$$s = \frac{14 + (s - 8) + (s - 6)}{2}$$

Simplifying this equation:

$$s = \frac{14 + 2s - 14}{2}$$ $$s = \frac{2s}{2}$$ $$s = s$$

This equation does not provide new information, so we use the fact that the radius of the incircle is 4 cm. The area of the triangle can be expressed as:

$$\text{Area} = r \cdot s = 4 \cdot s$$

Now, the area of the triangle can also be calculated using Heron's formula. For Heron's formula, the area is:

$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$

Substitute the known values for $a$, $b$, and $c$ into the formula to calculate the area. Equating the two expressions for the area will allow you to solve for $s$, and subsequently the values of $b$ and $c$.

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Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution:

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Let the quadrilateral be ABCD, and let the circle be the incircle of the quadrilateral, tangent to the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

Let O be the centre of the circle. We are required to prove that the opposite sides of the quadrilateral, namely AB and CD, and BC and DA, subtend supplementary angles at the centre O of the circle.

Consider the angles subtended by the sides AB and CD at the centre O. Let the angle subtended by side AB at O be $ \angle AOB $ and the angle subtended by side CD at O be $ \angle COD $.

Since the circle is the incircle of the quadrilateral, the tangents from a point to a circle are equal in length. Therefore, we have the following relationships:

  • AP = AS
  • BP = BQ
  • CQ = CR
  • DR = DS

Now, consider the two pairs of tangents from each vertex of the quadrilateral. From the properties of tangents, the tangents from a common external point to a circle subtend equal angles at the centre of the circle. Thus:

  • Angle $ \angle AOB = \angle COD $
  • Angle $ \angle BOC = \angle DOA $

Since the sum of angles around a point is $ 360^{\circ} $, we can write:

$ \angle AOB + \angle COD = 180^{\circ} $

Therefore, the opposite sides AB and CD subtend supplementary angles at the centre O.

Similarly, by considering the angles subtended by the sides BC and DA, we can prove that:

$ \angle BOC + \angle DOA = 180^{\circ} $

This completes the proof that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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