Two circles touch each other externally at C and AB is common tangent of circles, then is
70°
60°
100°
90°
(4)
Draw CM perpendicular to AB.

Now, AM = MC and MB = MC (tangents drawn from external point are equal).
⇒ AM = MC
⇒ = = 45°
(Since Δ AMC is right triangle)
∴ Also, MB = MC ⇒ = = 45° (Since Δ MBC is right angle triangle)
∴ = + = 45° + 45° = 90° ⇒ = 90°
If the circumference of a circle and the perimeter of a square are equal, then
Area of the circle = Area of the square
Area of the circle > Area of the square
Area of the circle < Area of the square
Nothing definite can be said about the relation between the areas of the circle and square.
(2)
Area of the circle > Area of the square
The area of the circle that can be inscribed in a square of 6cm is
(4)

ABCD is a square of side 6 cm. PQ is a diameter of the given circle such that
If two tangents inclined at an angle of are drawn to a circle of radius 3cm, then the length of each tangent is equal to
(4)

Angle between two tangents = (given)
Tangents are equally inclined to each other
and
(Tangent is perpendicular to the radius)
In
Hence, the length of each tangent is cm
In the given figure, tangents PA and PB to the circle centred at O, from point P are perpendicular to each other. If PA = 5 cm, then length of AB is equal to

cm
cm
cm
cm
(2)
PA = PB (Tangents from an external point P)
is a Right angle Triangle
In the given figure, PA and PB are two tangents drawn to the circle with centre O and radius 5 cm. If , then the length of PA is :

(2)
AB and CD are two chords of a circle intersecting at P. Choose the correct statement from the following:

(4)
[vertically opposite angle]
[Angle subtends on the same segment]
(by AA similarity)
In the given figure, AT is tangent to a circle centred at O. If , then is equal to

70°
50°
65°
40°
(4)
Given,
Maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is:
4
3
2
1
(3)
The maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is 2, as shown in

In Fig, O is the centre of the circle. MN is the chord and the tangent ML at point M makes an angle of 70° with MN. The measure of ∠MON is:

120°
140°
70°
90°
(2)
As ML is tangent at M and OM is radius of the circle, therefore ∠OML = 90°
⇒ ∠OMN + ∠NML = 90°
⇒ ∠OMN + 70° = 90° ⇒ ∠OMN = 20°
In ΔOMN, we have OM = ON (? radii of circle)
⇒ ∠OMN = ∠ONM = 20° (Angles opposite to equal sides are equal)
Also, ∠OMN + ∠MON + ∠ONM = 180°
⇒ 20° + ∠MON + 20° = 180°
⇒ ∠MON = 180° – 40° = 140°