OP Malhotra Class 10 Maths Solutions Chapter 14 Circle Exercise 14 (C)

OP Malhotra Class 10 Maths Solutions Chapter 14 Circle Ex 14(c)

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(c)

Question 1. (a) In figure, APB is tangent to circle with centre O. If ∠QPB = 50°, find ∠POQ.
Answer: In the given figure, APB is a line that touches the circle at point P. The circle's center is O. We are told that \( \angle QPB = 50° \). We know that OP is the radius of the circle and APB is the tangent at P. A key geometry rule states that the radius drawn to the point of tangency is perpendicular to the tangent. So, \( OP \perp APB \).
\( \implies \angle OPB = 90° \) This angle \( \angle OPB \) is made up of two smaller angles: \( \angle OPQ \) and \( \angle QPB \). So, \( \angle OPQ + \angle QPB = 90° \). We can substitute the given value: \( \angle OPQ + 50° = 90° \). To find \( \angle OPQ \), we subtract 50° from 90°.
\( \implies \angle OPQ = 90° - 50° \)
\( \implies \angle OPQ = 40° \) Now, consider the triangle \( \triangle OPQ \). In this triangle, OP and OQ are both radii of the same circle, so they must be equal in length. When two sides of a triangle are equal, the angles opposite those sides are also equal. This makes \( \triangle OPQ \) an isosceles triangle. Therefore, \( \angle OPQ = \angle OQP = 40° \). The sum of angles in any triangle is 180°. So, for \( \triangle OPQ \): \( \angle POQ + \angle OPQ + \angle OQP = 180° \) Substitute the known angle values: \( \angle POQ + 40° + 40° = 180° \). Combine the angles: \( \angle POQ + 80° = 180° \). Finally, subtract 80° from 180° to find \( \angle POQ \).
\( \implies \angle POQ = 180° - 80° \)
\( \implies \angle POQ = 100° \)In simple words: First, find the angle between the radius and the tangent, which is 90 degrees. Use this to get the angle inside the triangle. Since two sides of that triangle are radii, the angles opposite them are equal. Then, use the sum of angles in a triangle to find the final angle.

🎯 Exam Tip: Remember the property that the radius at the point of tangency is perpendicular to the tangent. This is a fundamental concept for solving such problems.

Question 1. (b) In figure, AB and AC are tangents. If AB = 4 cm, find AC.
Answer: In this problem, we have a circle, and from an external point A, two lines AB and AC are drawn that touch the circle. These lines are called tangents. A very important theorem in circle geometry states that the lengths of tangents drawn from an external point to a circle are equal. Here, AB and AC are both tangents drawn from the external point A to the circle. Since AB and AC are tangents from the same external point A, their lengths must be equal. We are given that \( AB = 4 \text{ cm} \). According to the tangent property, \( AC = AB \). Therefore, \( AC = 4 \text{ cm} \).In simple words: When you draw two lines from the same outside point and they both just touch a circle, those two lines will always be the same length. So, if one is 4 cm, the other must also be 4 cm.

🎯 Exam Tip: Always remember the theorem: "Lengths of tangents drawn from an external point to a circle are equal." This is a common and easy-to-apply rule.

Question 1. (c) In the figure, PQ and PR are tangents to circle, centre O. If ∠QPR = 80°, find ∠QOR.
Answer: In the given figure, PQ and PR are tangents to a circle with center O. We are given that \( \angle QPR = 80° \). When two tangents are drawn from an external point to a circle, the angle formed by these tangents (like \( \angle QPR \)) and the angle formed by the radii to the points of tangency (like \( \angle QOR \)) are supplementary. This means their sum is 180°. This happens because the radii OQ and OR are perpendicular to the tangents PQ and PR, respectively. So, \( \angle OQP = 90° \) and \( \angle ORP = 90° \). The quadrilateral OQPR has angles that sum to 360°. So, \( \angle QOR + \angle QPR = 180° \). We can substitute the given value: \( \angle QOR + 80° = 180° \). To find \( \angle QOR \), we subtract 80° from 180°.
\( \implies \angle QOR = 180° - 80° \)
\( \implies \angle QOR = 100° \)In simple words: When two lines touch a circle from an outside point, the angle between these lines and the angle formed by the center with the touch points always add up to 180 degrees. If one is 80 degrees, the other must be 100 degrees.

🎯 Exam Tip: A quadrilateral formed by two tangents from an external point and the two radii to the points of contact has two right angles, so the remaining two angles (tangent angle and central angle) are supplementary.

Question 2. In figure, O is the centre of the circle. Find ∠POS.
Answer: In the provided figure, O is the center of the circle. From an external point P, two tangents, PT and PS, are drawn to the circle. We are given that \( \angle TPO = 30° \). Consider the triangle \( \triangle PTO \). We know that the radius drawn to the point of tangency is perpendicular to the tangent. So, OT is the radius and PT is the tangent, which means \( OT \perp PT \). Therefore, \( \angle OTP = 90° \). In \( \triangle PTO \), the sum of all angles must be 180°. So, \( \angle TOP + \angle TPO + \angle OTP = 180° \). Substitute the known values: \( \angle TOP + 30° + 90° = 180° \). Combine the constant angles: \( \angle TOP + 120° = 180° \). To find \( \angle TOP \), subtract 120° from 180°.
\( \implies \angle TOP = 180° - 120° \)
\( \implies \angle TOP = 60° \) The line segment OP connects the external point P to the center O. This line OP bisects the angle formed by the two tangents (TPO) and also bisects the angle formed by the radii at the center (TOS). Since OP bisects \( \angle TOS \), it divides the angle into two equal halves. Thus, \( \angle TOP = \angle POS \). As we found \( \angle TOP = 60° \), it follows that \( \angle POS = 60° \).In simple words: First, find the missing angle inside the triangle formed by the tangent, radius, and line to the center, knowing one angle is 90 degrees. Then, remember that the line from the center to the outside point cuts the central angle into two equal parts, so the other angle is the same.

🎯 Exam Tip: Always recall that the line segment from the center to the external point from which tangents are drawn bisects the angle between the tangents and the angle between the radii to the points of contact.

Question 3. In figure, PQ is tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate
(i) ∠QAB
(ii) ∠PAD
(iii) ∠CDB
Answer:
(i) We are given that PQ is a tangent to the circle at point A, and AB is a chord of the circle. According to the alternate segment theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Here, \( \angle QAB \) is the angle between the tangent PQ and the chord AB. The angle in the alternate segment is \( \angle ADB \). We are given \( \angle ADB = 30° \). Therefore, \( \angle QAB = \angle ADB = 30° \).
In simple words: The angle that a tangent line makes with a chord is the same as the angle that the chord makes with any point on the opposite side of the circle. Since the angle ADB is 30 degrees, the angle QAB is also 30 degrees.

🎯 Exam Tip: The alternate segment theorem is crucial here: "The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment."

Answer:
(ii) We know that PQ is a straight line, which is a tangent. Angles on a straight line add up to 180°. The angles \( \angle PAD \), \( \angle DAB \), and \( \angle QAB \) lie on the straight line PQ. So, \( \angle PAD + \angle DAB + \angle QAB = 180° \). DB is given as a diameter. The angle subtended by a diameter at any point on the circumference is always a right angle (90°). Therefore, \( \angle DAB = 90° \) (angle in a semicircle). From part (i), we found \( \angle QAB = 30° \). Now, substitute these values into the equation: \( \angle PAD + 90° + 30° = 180° \). Combine the constant angles: \( \angle PAD + 120° = 180° \). To find \( \angle PAD \), subtract 120° from 180°.
\( \implies \angle PAD = 180° - 120° \)
\( \implies \angle PAD = 60° \)In simple words: On a straight line, all angles add up to 180 degrees. Since DB is a diameter, the angle it makes with point A on the circle is 90 degrees. We already found QAB. Now, subtract the known angles from 180 to find PAD.

🎯 Exam Tip: Remember two key facts: angles on a straight line sum to 180° and the angle in a semicircle is always 90°. These are essential for many circle geometry problems.

Answer:
(iii) Consider the triangle \( \triangle CDB \). We need to find \( \angle CDB \). We know that DB is the diameter, so the angle subtended by the diameter at point C on the circumference is \( \angle DCB = 90° \). This is an angle in a semicircle. We are given \( \angle CBD = 60° \). The sum of angles in any triangle is 180°. So, for \( \triangle CDB \): \( \angle CDB + \angle CBD + \angle DCB = 180° \). Substitute the known values: \( \angle CDB + 60° + 90° = 180° \). Combine the constant angles: \( \angle CDB + 150° = 180° \). To find \( \angle CDB \), subtract 150° from 180°.
\( \implies \angle CDB = 180° - 150° \)
\( \implies \angle CDB = 30° \)In simple words: In the triangle CDB, one angle (DCB) is 90 degrees because it's in a semicircle. Another angle (CBD) is given as 60 degrees. Since all angles in a triangle add up to 180 degrees, you can easily find the last angle (CDB) by subtracting the sum of the other two from 180.

🎯 Exam Tip: Always look for angles in a semicircle (which are 90°) and use the sum of angles in a triangle (180°) to find unknown angles.

Question 4. In figure, PR and PQ are the tangents, each of them being equal to 9 cm, ∠QPR = 60°. Find the length of the chord QR which joins their points of contact.
Answer: In the given figure, PQ and PR are two tangents drawn from an external point P to the circle. We are told that each tangent is 9 cm long, so \( PQ = PR = 9 \text{ cm} \). We are also given that the angle between these tangents, \( \angle QPR = 60° \). We need to find the length of the chord QR. Consider the triangle \( \triangle PQR \). We already know that \( PQ = PR = 9 \text{ cm} \). This means that \( \triangle PQR \) is an isosceles triangle because two of its sides are equal. In an isosceles triangle, the angles opposite the equal sides are also equal. So, \( \angle PQR = \angle PRQ \). The sum of angles in any triangle is 180°. For \( \triangle PQR \): \( \angle PQR + \angle PRQ + \angle QPR = 180° \). Substitute the given \( \angle QPR = 60° \): \( \angle PQR + \angle PRQ + 60° = 180° \). Since \( \angle PQR = \angle PRQ \), let's call them both \( x \). \( x + x + 60° = 180° \) \( 2x = 180° - 60° \) \( 2x = 120° \) \( x = \frac{120°}{2} \)
\( \implies x = 60° \) So, \( \angle PQR = \angle PRQ = 60° \). Now we have \( \angle QPR = 60° \), \( \angle PQR = 60° \), and \( \angle PRQ = 60° \). Since all three angles of \( \triangle PQR \) are 60°, it is an equilateral triangle. In an equilateral triangle, all sides are equal. Therefore, \( PQ = PR = QR \). Since \( PQ = 9 \text{ cm} \), then \( QR = 9 \text{ cm} \).In simple words: First, recognize that the two tangent lines are equal, making a special triangle. Use the sum of angles in a triangle to find the other two angles. If all angles are 60 degrees, it means all sides are equal, so the chord length is the same as the tangent length.

🎯 Exam Tip: When two tangents are drawn from an external point, the triangle formed by the external point and the two points of contact is always isosceles. If the angle at the external point is 60°, the triangle becomes equilateral.

Question 5. Find the length of the tangent drawn to a circle of radius 3 cm, from a point distant 5 cm from the centre.
Answer: Let O be the center of the circle. We are given that the radius of the circle is 3 cm. Let P be the external point from which the tangent is drawn. The distance of P from the center O is 5 cm. Let PQ be the tangent drawn from P to the circle, with Q being the point of contact. We know that the radius drawn to the point of tangency is perpendicular to the tangent. So, \( OQ \perp PQ \). This means that \( \triangle OQP \) is a right-angled triangle, with the right angle at Q ( \( \angle OQP = 90° \) ). We can use the Pythagorean theorem in \( \triangle OQP \), which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Here, OP is the hypotenuse, OQ is the radius, and PQ is the tangent length we need to find. \( OP^2 = OQ^2 + PQ^2 \) We are given: Radius \( OQ = 3 \text{ cm} \) Distance from center to point \( OP = 5 \text{ cm} \) Substitute these values into the Pythagorean theorem: \( (5)^2 = (3)^2 + PQ^2 \) Calculate the squares: \( 25 = 9 + PQ^2 \) To find \( PQ^2 \), subtract 9 from 25: \( PQ^2 = 25 - 9 \) \( PQ^2 = 16 \) Now, take the square root of 16 to find PQ:
\( \implies PQ = \sqrt{16} \)
\( \implies PQ = 4 \text{ cm} \) Since PQ and PR are tangents from the same external point P, their lengths are equal. So, \( PR = 4 \text{ cm} \). The length of the tangent is 4 cm.In simple words: Imagine a right triangle where one side is the circle's radius (3 cm), the long side is the distance from the center to the outside point (5 cm), and the third side is the tangent you need to find. Use the Pythagorean theorem to calculate the length of this third side.

🎯 Exam Tip: Always draw a diagram and recognize that the radius at the point of tangency forms a right angle with the tangent. This allows the application of the Pythagorean theorem directly.

Question 6. A circle touches the side BC of △ABC at P and touches AB and AC produced at Q and R respectively. If AQ = 5 cm, find the perimeter of △ABC.
Answer: Let the circle touch the side BC at P, and the extended sides AB and AC at Q and R respectively. O is the center of the circle. We are given that \( AQ = 5 \text{ cm} \). A key property of tangents from an external point to a circle is that their lengths are equal. 1. From external point A, tangents are drawn to the circle at Q and R. So, \( AQ = AR \). Since \( AQ = 5 \text{ cm} \), then \( AR = 5 \text{ cm} \). This is our first important equality. 2. From external point B, tangents are drawn to the circle at Q and P. So, \( BQ = BP \). This is our second important equality. 3. From external point C, tangents are drawn to the circle at P and R. So, \( CP = CR \). This is our third important equality. Now, let's find the perimeter of \( \triangle ABC \). The perimeter is the sum of the lengths of its sides: Perimeter of \( \triangle ABC = AB + BC + AC \) We can break down BC into BP and CP: Perimeter of \( \triangle ABC = AB + (BP + CP) + AC \) Now, substitute the equalities from the tangent properties: Replace BP with BQ (from point B) and CP with CR (from point C): Perimeter of \( \triangle ABC = AB + BQ + CR + AC \) Rearrange the terms to group them differently: Perimeter of \( \triangle ABC = (AB + BQ) + (AC + CR) \) Notice that \( AB + BQ \) is the total length of the tangent \( AQ \). And \( AC + CR \) is the total length of the tangent \( AR \). So, Perimeter of \( \triangle ABC = AQ + AR \) We know that \( AQ = 5 \text{ cm} \) and \( AR = 5 \text{ cm} \). Perimeter of \( \triangle ABC = 5 \text{ cm} + 5 \text{ cm} \) Perimeter of \( \triangle ABC = 10 \text{ cm} \)In simple words: The total distance around the triangle can be found by adding up the two tangent lines that start from point A. This is because the other parts of the triangle's perimeter can be swapped with parts of those main tangent lines. If one tangent is 5 cm, the other is also 5 cm, so the total perimeter is 10 cm.

🎯 Exam Tip: Remember that the perimeter of the triangle in this configuration is always twice the length of the tangent from the external vertex (A) to the circle. This shortcut can save time.

Question 7. There are two concentric circles of radii 3 cm and 5 cm respectively. Find the length of the chord of the outer circle which touches the inner circle.
Answer: We have two circles that share the same center, O. These are called concentric circles. The radius of the larger circle is 5 cm. Let's call a point on its circumference A, so \( OA = 5 \text{ cm} \). The radius of the smaller circle is 3 cm. Let's call a point on its circumference P, so \( OP = 3 \text{ cm} \). We are looking for the length of a chord of the larger circle that touches the smaller circle. Let this chord be AB, and let it touch the smaller circle at point P. When a chord of the larger circle touches the smaller circle at P, it means the chord AB is tangent to the smaller circle at P. We know that the radius drawn to the point of tangency is perpendicular to the tangent. So, \( OP \perp AB \). This means that \( \triangle OPA \) is a right-angled triangle, with the right angle at P ( \( \angle OPA = 90° \) ). We can use the Pythagorean theorem in \( \triangle OPA \): \( OA^2 = OP^2 + AP^2 \) Substitute the known radii: \( (5)^2 = (3)^2 + AP^2 \) Calculate the squares: \( 25 = 9 + AP^2 \) To find \( AP^2 \), subtract 9 from 25: \( AP^2 = 25 - 9 \) \( AP^2 = 16 \) Now, take the square root of 16 to find AP:
\( \implies AP = \sqrt{16} \)
\( \implies AP = 4 \text{ cm} \) Since OP is perpendicular to the chord AB, it bisects the chord AB. This means AP = PB. So, the total length of the chord AB is \( 2 \times AP \). \( AB = 2 \times 4 \text{ cm} \)
\( \implies AB = 8 \text{ cm} \) The length of the chord of the outer circle that touches the inner circle is 8 cm.In simple words: First, draw a right-angled triangle using the radii of both circles and half of the chord. Use the Pythagorean theorem to find the length of half the chord. Then, double this length to get the full chord.

🎯 Exam Tip: Always remember that a radius drawn to a chord that is also tangent to an inner concentric circle will be perpendicular to that chord and bisect it. This creates a right-angled triangle for Pythagorean calculations.

Question 8. Three circles with centres A, B, C touch each other externally; AB = 4 cm, BC = 6 cm, CA = 7 cm; find their radii.
Answer: Let the radii of the three circles with centers A, B, and C be \( x \), \( y \), and \( z \) respectively. When two circles touch each other externally, the distance between their centers is equal to the sum of their radii. We are given the distances between the centers: 1. Distance between center A and center B is \( AB = 4 \text{ cm} \). So, \( x + y = 4 \) cm. (Equation 1) 2. Distance between center B and center C is \( BC = 6 \text{ cm} \). So, \( y + z = 6 \) cm. (Equation 2) 3. Distance between center C and center A is \( CA = 7 \text{ cm} \). So, \( z + x = 7 \) cm. (Equation 3) Now we have a system of three linear equations. To solve for \( x, y, z \), we can add all three equations together: \( (x + y) + (y + z) + (z + x) = 4 + 6 + 7 \) Combine like terms: \( 2x + 2y + 2z = 17 \) Factor out 2: \( 2(x + y + z) = 17 \) Divide by 2 to find the sum of the radii: \( x + y + z = \frac{17}{2} \)
\( \implies x + y + z = 8.5 \text{ cm} \). (Equation 4) Now, we can find each radius by subtracting one of the initial equations from Equation 4: To find \( z \): Subtract Equation 1 ( \( x + y = 4 \) ) from Equation 4 ( \( x + y + z = 8.5 \) ). \( (x + y + z) - (x + y) = 8.5 - 4 \) \( z = 4.5 \text{ cm} \) To find \( x \): Subtract Equation 2 ( \( y + z = 6 \) ) from Equation 4 ( \( x + y + z = 8.5 \) ). \( (x + y + z) - (y + z) = 8.5 - 6 \) \( x = 2.5 \text{ cm} \) To find \( y \): Subtract Equation 3 ( \( z + x = 7 \) ) from Equation 4 ( \( x + y + z = 8.5 \) ). \( (x + y + z) - (z + x) = 8.5 - 7 \) \( y = 1.5 \text{ cm} \) So, the radii of the three circles are: Radius of circle A (\( x \)) = 2.5 cm Radius of circle B (\( y \)) = 1.5 cm Radius of circle C (\( z \)) = 4.5 cmIn simple words: When circles touch each other on the outside, the distance between their centers is just the sum of their radii. Set up three equations for the three circles. Add them all up to find the sum of all radii. Then, subtract each original equation from this sum to find each individual radius.

🎯 Exam Tip: Remember the rule for externally touching circles: the distance between their centers is the sum of their radii. Solving this becomes a system of linear equations, a common algebra technique.

Question 9. Equal circles, centres O and O' touch each other at X. OO' is produced to meet the circle O' at A. AC is tangent to the circle whose centre is O. O' is perpendicular to AC. Find the value of
(i) \( \frac{AO'}{AO} \)
(ii) \( \frac{\text{area of } \triangle ADO'}{\text{area of } \triangle ACO} \)
Answer: Let \( r \) be the radius of each of the equal circles, with centers O and O'. They touch externally at X. So, \( OX = O'X = r \). The line segment OO' goes through X, so \( OO' = OX + O'X = r + r = 2r \). The line OO' is extended to meet circle O' at A. This means A is a point on the circumference of circle O', and A is on the line extending from O through O'. So, \( AO' \) is a radius of circle O', which means \( AO' = r \). Now, let's find the length of AO. \( AO = AX + XO \) Since \( A \) is on the line O'OX extended, \( A \) is the point where the line O'O touches the circle O' on the left. So, \( AO' = r \). \( AO = AX + XO \) \( AX = AO' + O'X = r + r = 2r \). Wait, this is not correct. Let's re-evaluate A's position: OO' is produced to meet circle O' at A. This means A is on the line segment OO' extended *past* O', so it's on the left side of O'. So, \( A \) is such that \( AO' = r \). Then the distance from O to A would be \( AO = OO' + O'A \). No, this is incorrect based on the diagram. From the diagram, A is on the far left, then O', then X, then O. So, \( AO = AO' + O'X + XO \). Since \( AO' \) is the radius of circle O' (touching at A), \( AO' = r \). \( O'X \) is the radius of circle O', so \( O'X = r \). \( XO \) is the radius of circle O, so \( XO = r \). Thus, \( AO = r + r + r = 3r \). We are given that AC is tangent to the circle with center O. O'D is perpendicular to AC. Consider \( \triangle ADO' \) and \( \triangle ACO \). We know that \( AC \) is tangent to circle O at C. Therefore, the radius OC is perpendicular to the tangent AC. So, \( \angle ACO = 90° \). We are also given that \( O'D \perp AC \). So, \( \angle ADO' = 90° \). Both triangles have a common angle at A, so \( \angle A = \angle A \) (common angle). Since two angles of \( \triangle ADO' \) are equal to two angles of \( \triangle ACO \), the triangles are similar by AA (Angle-Angle) axiom. \( \triangle ADO' \sim \triangle ACO \) (i) Since the triangles are similar, the ratio of their corresponding sides is equal. \( \frac{AO'}{AO} = \frac{DO'}{CO} = \frac{AD}{AC} \) We need to find \( \frac{AO'}{AO} \). We established that \( AO' = r \) (radius of circle O'). And \( AO = 3r \) (distance from A to O, covering the three radii \( AO', O'X, XO \)). Therefore, \( \frac{AO'}{AO} = \frac{r}{3r} = \frac{1}{3} \).In simple words: The ratio of distances from point A to the centers O' and O is needed. First, figure out all distances in terms of 'r', the radius of one circle. Since the triangles formed are similar, the ratio of their matching sides will be the same.

🎯 Exam Tip: When circles touch, the line connecting their centers passes through the point of contact. This allows you to express distances in terms of radii. Also, look for similar triangles using AA similarity criterion (common angle and right angles).

Answer:
(ii) For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Since \( \triangle ADO' \sim \triangle ACO \), \( \frac{\text{area of } \triangle ADO'}{\text{area of } \triangle ACO} = \left(\frac{AO'}{AO}\right)^2 \) From part (i), we found \( \frac{AO'}{AO} = \frac{1}{3} \). Substitute this value into the area ratio: \( \frac{\text{area of } \triangle ADO'}{\text{area of } \triangle ACO} = \left(\frac{1}{3}\right)^2 \)
\( \implies \frac{\text{area of } \triangle ADO'}{\text{area of } \triangle ACO} = \frac{1^2}{3^2} \)
\( \implies \frac{\text{area of } \triangle ADO'}{\text{area of } \triangle ACO} = \frac{1}{9} \)In simple words: Since the two triangles are similar, the ratio of their areas is simply the square of the ratio of their sides. We found the side ratio to be 1/3, so squaring it gives 1/9 for the area ratio.

🎯 Exam Tip: Remember the theorem: "The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides." This is a key property of similar figures.

Question 10. In figure, P and Q are centres of two circles of radii 12 cm and 3 cm respectively. A and B are the points of contact of the common tangent XY. Find AB.
Answer: We have two circles with centers P and Q. The radius of circle P is \( PA = 12 \text{ cm} \), and the radius of circle Q is \( QB = 3 \text{ cm} \). The circles touch externally. XY is a common tangent, touching circle P at A and circle Q at B. We need to find the length of AB. 1. **Draw radii and perpendiculars:** Draw radii PA and QB to the points of contact A and B respectively. Since the tangent XY is perpendicular to the radii at the points of contact, we have: \( PA \perp XY \) \( QB \perp XY \) This means \( \angle PAX = 90° \) and \( \angle QBY = 90° \). 2. **Distance between centers:** Since the two circles touch externally, the distance between their centers P and Q is the sum of their radii. \( PQ = PA + QB = 12 \text{ cm} + 3 \text{ cm} = 15 \text{ cm} \). 3. **Construct a rectangle/right triangle:** Draw a line segment QS parallel to AB, from Q to PA. Let S be the point where this line meets PA. Since PA and QB are both perpendicular to XY, PA is parallel to QB. With QS parallel to AB, the figure ABSQ forms a rectangle. Therefore, \( AB = SQ \). Also, \( SA = QB = 3 \text{ cm} \). 4. **Find PS:** The length PS is the difference between PA and SA. \( PS = PA - SA = 12 \text{ cm} - 3 \text{ cm} = 9 \text{ cm} \). 5. **Apply Pythagorean theorem:** Now consider the right-angled triangle \( \triangle PSQ \). It is a right-angled triangle because QS is parallel to AB, and PA is perpendicular to AB, making QS perpendicular to PS. In \( \triangle PSQ \), we have: Hypotenuse \( PQ = 15 \text{ cm} \) Side \( PS = 9 \text{ cm} \) Side \( SQ \) (which is equal to AB) needs to be found. Using the Pythagorean theorem: \( PQ^2 = PS^2 + SQ^2 \) Substitute the known values: \( (15)^2 = (9)^2 + SQ^2 \) Calculate the squares: \( 225 = 81 + SQ^2 \) To find \( SQ^2 \), subtract 81 from 225: \( SQ^2 = 225 - 81 \) \( SQ^2 = 144 \) Now, take the square root of 144 to find SQ:
\( \implies SQ = \sqrt{144} \)
\( \implies SQ = 12 \text{ cm} \) Since \( AB = SQ \), the length of the common tangent AB is 12 cm.In simple words: First, draw lines from the centers to the tangent points and a line connecting the centers. Create a right-angled triangle by drawing a line from the smaller radius's center parallel to the tangent. Use the Pythagorean theorem with the distance between centers and the difference in radii to find the length of the parallel line, which is the same as the tangent length.

🎯 Exam Tip: For common tangents between two circles, construct a right-angled trapezoid and then a right-angled triangle by drawing a line parallel to the tangent through the center of the smaller circle. This setup allows the use of the Pythagorean theorem effectively.