10th Grade > Mathematics
CIRCLES MCQs
:
At most one circle can be drawn through a given set of three distinct points. These three points will then be referred to as 'concyclic points' (Lying on the same circle). .
:
A
We have
OA ⊥ AP and OB ⊥ BP [ The tangent at any point of a circle is perpendicular to the radius through the point of contact].
Join OP.
In right Δ OAP, we have
OA = 8 cm, AP = 6 cm
∴ OP2=OA2+AP2 [by Pythagoras theorem]
⇒ OP=√OA2+AP2=√82+62cm=√100cm=10 cm
In right Δ OBP, we have
OB = 6 cm, OP = 10 cm
∴ OP2=OB2+BP2
[by Pythagoras' theorem]
⇒ BP=√OP2−OB2=√102−62cm=√64cm
Thus, the length of BP
=√64cm = 8 cm.
:
A, B, C, and D
Given that AC = BD.
OA = OB (radius)
Adding the above equations,
AC + OA = BD + OB
⟹OC = OD
⟹∠ODC=∠OCD (Property of isosceles triangles)
Given ∠COD=90∘
Therefore, ∠ODC=∠OCD=45∘ (Base angles are equal)
Join OP.
Since a tangent at any point of a circle is perpendicular to the radius at the point of contact, we have OP ⊥ CD.
Consider right angled triangle ODP,
tan ∠ODP=OPPD
Tan45∘=100PD (∵OP=1m=100cm)
⟹1=100PD
⇒PD=100 cm
Similarly we can show that PC = 100 cm
Now, sin∠ODP=OPOD
⟹sin45∘=0.7071=100OD
i.e., OD=1000.7071
⇒OD=141.42 cm
But, BD=OD−OB
⇒BD=141.42–100=41.42 cm
AC+CP=BD+100 (since AC = BD and PC= 100 cm)
=41.42+100=141.42cm
:
C
In ΔAOC,
OA=OC --------(radii of the same circle)
∴ΔAOC is an isosceles triangle
→∠OAC=∠OCA=25∘----- (base angles of an isosceles triangle )
In ΔBOC,
OB=OC --------(radii of the same circle)
∴ΔBOC is an isosceles triangle
→∠OBC=∠OCB=35∘ -----(base angles of an isosceles triangle )
∠ACB=25∘+35∘=60∘
∠AOB=2×∠ACB ----(angle at the center is twice the angle at the circumference)
= 2×60∘
=120∘
:
A line that touches a circle at only one point is called a tangent.