Two equal circles cut at X and Y. If C1, and C2, are the centres of the circles, prove that XY is the right bisector of C1C2
Answers
Step-by-step explanation:
Given :-
Two equal circles cut at X and Y.
C1, and C2, are the centres of the circles.
Required To prove :-
XY is the right bisector of C1 C2
Construction:-
Join C1 X and C1 Y , C2 X and C2 Y
Proof :-
Given that
C1 and C2 are the centres of the two circles.
In ∆ C1 X C2 and ∆ C1 Y C2
C1 X ≅ C1 Y ( Radii of the circles )
C2 X ≅ C2 Y ( Radii of the circles )
C1 C2 ≅ C1 C2 ( Common sides )
By SSS Theorem of Congruence
∆ C1 X C2 ≅ ∆ C1 Y C2
=> ∠ X C1 C2 ≅ ∠ Y C1 C2
Since, Corresponding parts in the Congruent triangles are congruent
In ∆ X C1 O and ∆ Y C1 O
X C1 ≅ Y C1 ( Radius )
∠X C1 C2 ≅ ∠Y C1 C2
C1 O ≅ C1 O ( common side)
By SAS theorem of Congruence
∆ X C1 O ≅ ∆ Y C1 O
XO ≅ OY
Since, Corresponding parts in the Congruent triangles are congruent
∠C1 O X ≅ ∠C1 O Y
=> ∠ C1 O X + ∠ C1 O Y = 180° ( linear pair)
=> 2 ∠C1 O X = 2∠ C1 O Y = 180°
=> ∠ C1 O X = ∠C1 O Y = 90°
Therefore, XY is the perpendicular bisector of C1 C2 .
Hence, Proved.
CONCEPT :
- We have to first draw a tangent to he circle and then prove it as shown below.
- The tangent of a circle at any point in a circle is perpendicular to the radius through the point of contact.
information about Circle :
- A circle is a shape consisting of all the points in a plane which are at a fixed distance (radius) from a given point(centre).
- Circle is the locus of points equidistant from a given point, the center of the circle. The common distance from the center of the circle to its points is called radius.
QUESTION :
- Two equal circles cut at X and Y. If C1, and C2, are the centres of the circles, prove that XY is the right bisector of C1C2
EXPLANATION :
C1 X = C, Y
O C 1 = O C 1 ( common )
< Y X C 1 = < X Y C1
in traingle
∆ X C1 O = ∆ C1 O Y
< X C1 Y = 90 °
< C1 X Y = 45° = < C1 Y X
similarly
∆ C2 O X = ∆ C2 O Y
< C2 X Y = C2 Y X = 45 °
hence, O C2 X = O C2 Y
hence, X Y in it bisector of
C1 C2
