Question

two circles of radius 10 cm and 8cm intersect each other and the length of the chord is 12cm find the distance between the centres

Answer 1

• enter of circles= O and O'
• Radii = 10 and 8cm
• common chord= PQ
• OP= 10cm
• O'P = 8cm
• PQ= 12cm
• PL= 1/2PQ
• = 1/2 x 12
• = 6cm
• In right angle triangle OLP 
• OP² = OL² + LP²
• OL= √OP² - LP² 
• = √10²- 6²
• = √64
• = 8cm
• In right angle triangle O'LP
• O'P²= O'L² - LP²
• O'L= √O'P² - LP²
• = √8² - 6²
• = √28
• = 5.29 cm
• OO'= 8 + 5.29 
• = 13.29 cm

Answer 2

ANSWER:

______________________________

Length of the common chord AB = 12 cm.

Let the radius of the circle with centre O  is OA= 10 cm.

Radius of circle with centre P  is AP=8 cm

From the figure, 
OP perpendicular to AB.

$= > AC = CB$
Therefore,

AC =6 cm.........(since AB =12 cm)

In Δ ACP,

${AP}^{2} = {PC}^{2} + {AC}^{2} ...........(by \: pythagoras \: theorem)$

${8}^{2} = {PC}^{2} + {6}^{2} \\ \\ {PC}^{2} = 64 - 36 = 28cm$

$PC = 2 \sqrt{7}$
Now,

Consider Δ ACO,

${AO}^{2} = {OC}^{2} + {AC}^{2} ........(by \: pythagoras \: theorem)$

${10}^{2} = {OC}^{2} + {6}^{2}$
${OC}^{2} = 100 - 36$
$OC ^{2} = 64$
$OC = 8cm.....taking \: sqrt \: on \: both \: the \: sides$
From the figure,  

OP= OC +PC =8+2√7cm.

Hence the distance between the center is
(8+2√7)cm