Question

21. Two equal circles of radius r intersect such that each pass through the
centre of the other. The length of the common chord of the circle is

Answer 1

Answer:

let the circles intersect at 2 point say A,B and centres of circles be C and D

the common chord is AB and it is perpendicular bisector of CD let intersection point of AB AND CD be P

now,CP=PD=r/2 and APC=90

and AC =r as A is a point on circle

as APC =90,IN TRIANGLE APC

by pythagoras theroem

Ap^2+PC^2=AC^2

r^2=(r/2)^2+X^2

X^2=3r^2/4

$4\sqrt{3}$

Answer 2

Answer:

let the circles intersect at 2 point say A,B and centres of circles be C and D

the common chord is AB and it is perpendicular bisector of CD let intersection point of AB AND CD be P

now,CP=PD=r/2 and APC=90

and AC =r as A is a point on circle

as APC =90,IN TRIANGLE APC

by pythagoras theroem

Ap^2+PC^2=AC^2

r^2=(r/2)^2+X^2

X^2=3r^2/4

4\sqrt{3}4

3

Step-by-step explanation:

hope its help you