CBSE 10th Standard Maths Subject Circles HOT Questions 2 Mark Questions With Solution 2021
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CBSE 10th Standard Maths Subject Circles HOT Questions 2 Mark Questions With Solution 2021
10th Standard CBSE

Reg.No. :
Maths

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6 cm respectively. Find the sides AB and AC.
(a) 
In the given figure, AB is a diameter of the circle, with centre O and AT is a tangent. Calculate the numerical value of x.
(a) 
In given figure, find the perimeter of \(\angle ABC\), if AP = 10 cm.
(a) 
Two circles intersect each other at two points A and B. From point A, tangents AP and AQ are drawn to two circles which intersect the circles at the pointsP and Q respectively. Prove that AB is the bisector of \(\angle PBQ\).
(a) 
Let A be a point of intersection of two intersecting circles with centres O and O'. The tangents at A to the two circles meet the circles at B and C respectively. Point P is located so that AOPO; is a parallelogram. Prove that P is the circumcentre of \(\triangle ABC\).
(a)
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CBSE 10th Standard Maths Subject Circles HOT Questions 2 Mark Questions With Solution 2021 Answer Keys

13 cm, 15 cm

\(x={ 58 }^{ \circ }\)

20 cm

Let O and O' be the centres of two circles, tangents AP and AQ intersect these circles in A and the circle with centre O in P and the circle with centre O' in Q. Join PBQ and AB.
Now, AP is a tangent to the circle with centre O' and AB is the chord.
ஃ BAP = AQB .......(i)
[ s in the corresponding alternate segment]
Again, AQ is a tangent to the circle with centre O and AB is the chord.
ஃ \(\angle \)BAQ = \(\angle \)BPA .....(ii)
Now, by angle sum property, we have
\(\angle \)BPA + \(\angle \)BAP + \(\angle \)ABP = 180°
\(\angle \)AQB + \(\angle \)BAQ + \(\angle \)QBA = 180°
FRom (iii) and (iv), we have
\(\angle \)BPA + \(\angle \)BPA + \(\angle \)ABP = \(\angle \)AQB + \(\angle \)BAQ + \(\angle \)QBA
⇒ \(\angle \)ABP = \(\angle \)QBA [using (i) & (ii)]
hence, AB is the bisector of \(\angle \)PBQ. 
Since AB and AC are two tangents from Point A to the two circle.
ஃ OA ⊥ AC and O'A ⊥ AB
[∵ radius through the point of contact is perpendicular to the tangents]
Let OP intersects AB at M, therefore OM ⊥ AB
ஃ AM = BM
[∵ ⊥ from the centre of a circle to a chord bisects the chord]
ஃ OM and hence OP is the perpendicular bisector of AB.
Similarly PO' is perpendicular of AC.
Now, in \(\Delta \)ABC
OP is the perpendicular bisector of side AB.
ஃ PA = PB
[∵ any point on the perpendicular bisector is equidistant from the fixed point]
Similarly, PA = PC
Hence, PA = PB = PC
ஃ P is equidistant from the three vertices of \(\Delta \)ABC.
Hence, P is the circumcentre of \(\Delta \)ABC.
As, AOPO' is a parallelogram.
ஃ AO  O'P [opposite sides of a  gm]
ஃ OP ⊥ AB
[ ∵ AO'  OP and O'A ⊥ AB, ஃ OP ⊥ AB]