Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. find the length of the common chord.

Transcript Ex 9.2, 1 Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. Given: Circle C1 with radii 5cm & C2 with radii 3cm Intersecting at P & Q. OP = 5cm , XP = 3cm & OX = 4cm To find: Length of common chord i.e., length of PQ Solution: Let the point where OX intersects PQ be R. In Δ POX & Δ QOX OP = OQ XP = XQ OX = OX ∴ Δ POX ≅ Δ QOX ∠ POX = ∠ QOX Also, In Δ POR & Δ QOR OP = OQ ∠ POR = ∠ QOR OR = OR ∴ Δ POR ≅ Δ QOR ⇒ ∠ PRO = ∠ QRO & PR = RQ Since PQ is a line ∠ PRO + ∠ QRO = 180° ∠ PRO + ∠ PRO = 180° 2∠ PRO = 180° ∠ PRO = (180°)/2 ∠ PRO = 90° Therefore, ∠ QRO = ∠ PRO = 90° Also, ∠ PRX = ∠ QRO = 90° Let OR = x, So, XR = OX – OR = 4 – x Now, From (4) & (5) 52 – x2 = –7 – x2 + 8x 25 – x2 = –7 – x2 + 8x 25 + 7 – x2 + x2 = 8x 32 = 8x 8x = 32 x = 32/8 x = 4 Putting value of x in (4) PR2 = 25 – x2 PR2 = 25 – 42 PR2 = 25 – 16 PR2 = 9 PR = √9 = 3 ∴ PQ = 2PR = 2 × 3 = 6 Hence, length of common chord = 6 m Note: OR = x = 4 cm & XR = 4 – x = 4 – 4 = 0 cm since XR = 0 this means point X & R coincide Hence actual figure is as follows Show More

• Engineering and Architecture • Computer Application and IT • Pharmacy • Hospitality and Tourism • Competition • School • Study Abroad • Arts, Commerce & Sciences • Management and Business Administration • Learn • Online Courses and Certifications • Medicine and Allied Sciences • Law • Animation and Design • Media, Mass Communication and Journalism • Finance & Accounts Given:Two circles of radii and intersect at two points and the distance between their centres is . To find the length of the common chord. Construction: Join OP and draw Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB. Let, PM = x , so QM=4-x In APM, using Pythagoras theorem ...........................1 Also, In AQM, using Pythagoras theorem ...........................2 From 1 and 2, we get Put,x=0 in equation 1

From the following figure, AP = 7 cm, PB = 5 cm, and AB = 10 cm Let AE = x cm, so EB = (10 – x) cm The length of the common chord PQ In right angled ΔAPE, AP 2 = PE 2 + AE 2 PE 2 = 7 2– x 2–––(1) In right angled ΔPBE, PB 2 = PE 2 + BE 2 PE 2 = 5 2– (10 – x) 2 PE 2 = 25 – 100 – x 2 + 20x ------ (2) From equation (1) and equation (2) 7 2– x 2 = 25 – 100 – x 2 + 20x ⇒ 20x = 100 + 49 – 25 ⇒ x = 124/20 ⇒ x = 31/5 From equation (1) PE 2 = 49 – (31/5) 2 = 49 – 961/25 ⇒ PE 2 = (1225 – 961)/25 = 264/25 ⇒ PE = √264/25 ⇒ PE = 2√66/5 As we know, PQ = 2PE ⇒ PQ = 2 × 2√66/5 ⇒ PQ = 4√66/5

Q5.Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a hall to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?

Two circles having radii $7$cm & $19$ cm are separated by a distance of $22$ cm between their centers. If they are intersecting each other at two points $P$& $Q$ then what will be the length of the common chord PQ? a.) $\frac$ I tried this by assuming one circle to be centered at the origin but it's creating utter mess. I am 12th grade. Thanks for your help! OK, i found a $r_1, r_2$ separated by a distance $d$ between their center $$=\frac$$

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Q L = x t h e n O ′ L = ( 4 − x ) P Q = y a n d P L = 2 y ​ I n Δ O L P , O L 2 = O P 2 − L P 2 = 5 2 − ( 2 y ​ ) 2 ⇒ x 2 = 2 5 − 4 y 2 ​ − − − − − ( 1 ) I n Δ O ′ L P , O ′ L 2 = ( 3 2 − 4 y 2 ​ ) ( 4 − x ) 2 = 9 − 4 y 2 ​ − − − − − ( 2 ) f r o m ( 1 ) a n d ( 2 ) w e g e t x 2 − 1 6 + 8 x − x 2 = 2 5 − 4 y 2 ​ − 9 + 4 y 2 ​ ⇒ − 1 6 + 8 x = 1 6 ⇒ 8 x = 3 2 ∴ x = 4 f r o m e q u n ( 2 ) 1 6 = 2 5 − 4 y 2 ​ ⇒ − 9 = − 4 y 2 ​ ⇒ y 2 = 3 6 ∴ y = 6 c m ​

Transcript Ex 9.2, 1 Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. Given: Circle C1 with radii 5cm & C2 with radii 3cm Intersecting at P & Q. OP = 5cm , XP = 3cm & OX = 4cm To find: Length of common chord i.e., length of PQ Solution: Let the point where OX intersects PQ be R. In Δ POX & Δ QOX OP = OQ XP = XQ OX = OX ∴ Δ POX ≅ Δ QOX ∠ POX = ∠ QOX Also, In Δ POR & Δ QOR OP = OQ ∠ POR = ∠ QOR OR = OR ∴ Δ POR ≅ Δ QOR ⇒ ∠ PRO = ∠ QRO & PR = RQ Since PQ is a line ∠ PRO + ∠ QRO = 180° ∠ PRO + ∠ PRO = 180° 2∠ PRO = 180° ∠ PRO = (180°)/2 ∠ PRO = 90° Therefore, ∠ QRO = ∠ PRO = 90° Also, ∠ PRX = ∠ QRO = 90° Let OR = x, So, XR = OX – OR = 4 – x Now, From (4) & (5) 52 – x2 = –7 – x2 + 8x 25 – x2 = –7 – x2 + 8x 25 + 7 – x2 + x2 = 8x 32 = 8x 8x = 32 x = 32/8 x = 4 Putting value of x in (4) PR2 = 25 – x2 PR2 = 25 – 42 PR2 = 25 – 16 PR2 = 9 PR = √9 = 3 ∴ PQ = 2PR = 2 × 3 = 6 Hence, length of common chord = 6 m Note: OR = x = 4 cm & XR = 4 – x = 4 – 4 = 0 cm since XR = 0 this means point X & R coincide Hence actual figure is as follows Show More

• Engineering and Architecture • Computer Application and IT • Pharmacy • Hospitality and Tourism • Competition • School • Study Abroad • Arts, Commerce & Sciences • Management and Business Administration • Learn • Online Courses and Certifications • Medicine and Allied Sciences • Law • Animation and Design • Media, Mass Communication and Journalism • Finance & Accounts Given:Two circles of radii and intersect at two points and the distance between their centres is . To find the length of the common chord. Construction: Join OP and draw Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB. Let, PM = x , so QM=4-x In APM, using Pythagoras theorem ...........................1 Also, In AQM, using Pythagoras theorem ...........................2 From 1 and 2, we get Put,x=0 in equation 1

Q L = x t h e n O ′ L = ( 4 − x ) P Q = y a n d P L = 2 y ​ I n Δ O L P , O L 2 = O P 2 − L P 2 = 5 2 − ( 2 y ​ ) 2 ⇒ x 2 = 2 5 − 4 y 2 ​ − − − − − ( 1 ) I n Δ O ′ L P , O ′ L 2 = ( 3 2 − 4 y 2 ​ ) ( 4 − x ) 2 = 9 − 4 y 2 ​ − − − − − ( 2 ) f r o m ( 1 ) a n d ( 2 ) w e g e t x 2 − 1 6 + 8 x − x 2 = 2 5 − 4 y 2 ​ − 9 + 4 y 2 ​ ⇒ − 1 6 + 8 x = 1 6 ⇒ 8 x = 3 2 ∴ x = 4 f r o m e q u n ( 2 ) 1 6 = 2 5 − 4 y 2 ​ ⇒ − 9 = − 4 y 2 ​ ⇒ y 2 = 3 6 ∴ y = 6 c m ​