in the adjoining figure, two equal circles with centre O and O' touche each other at X OO' produced meet the circle with centre O' at A . AC is tangent to the circle with centre O at the point C. O'D is perpendicular to AC find the value of DO'/CO is
PLEASE GIVE CORRECT ANSWER
PLEASE GIVE CORRECT ANSWER
Answers & Comments
Answer:
search-icon-header
Search for questions & chapters
search-icon-image
Question
Bookmark
In the figure, two equal circles, with centres O and O
′
, touch each other at X.O
′
X produced meets the circle with centre O at A.AC is tangent to the circle with centre O, at the point C.
O
′
D is perpendicular to AC. Find the value of
CO
DO
′
.
494161
expand
Medium
Solution
verified
Verified by Toppr
O
′
D is perpendicular to AC.
We know that ∠ADO
′
=90
o
Radius OC is perpendicular to tangent AC.
In ΔADO
′
and ΔACO,
∠ADO
′
=∠ACO (each 90 degree)
∠DAO=∠CAO (common)
By AA property, trangles ADO' and ACO are similar to each other.
AO
AO
′
=
CO
DO
′
(corresponding sides of similar triangles)
AO=AO
′
+O
′
X+OX
=3AO
′
(Since AO
′
=O
′
X=OX because radii of the two circles are equal)
AO
AO
′
=
3AO
AO
′
=
3
1
CO
DO
′
=
AO
AO
′
=
3
1
CO
DO
′
=
3
1
.
Verified answer
Step-by-step explanat
In the figure, two equal circles, with centres O and O
′
, touch each other at X.O
′
X produced meets the circle with centre O at A.AC is tangent to the circle with centre O, at the point C.
O
′
D is perpendicular to AC. Find the value of
CO
DO
′
.
494161
expand
Medium
Solution
verified
Verified by Toppr
O
′
D is perpendicular to AC.
We know that ∠ADO
′
=90
o
Radius OC is perpendicular to tangent AC.
In ΔADO
′
and ΔACO,
∠ADO
′
=∠ACO (each 90 degree)
∠DAO=∠CAO (common)
By AA property, trangles ADO' and ACO are similar to each other.
AO
AO
′
=
CO
DO
′
(corresponding sides of similar triangles)
AO=AO
′
+O
′
X+OX
=3AO
′
(Since AO
′
=O
′
X=OX because radii of the two circles are equal)
AO
AO
′
=
3AO
AO
′
=
3
1
CO
DO
′
=
AO
AO
′
=
3
1
CO
DO
′
=
3
1
.