Plane Geometry
- Two circles are of radii 7 cm and 2 cm their centres being 13cm apart. Then the length of direct common tangent to the circles between the points of contact is
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As per the given in question , we draw a figure of two circles with radii 7 cm and 2 cm ,
As we know that , Direct common tangent = √(Distance between two centres)² - (r1 - r2)²
Direct common tangent = √(C1C2)² -(r1 - r2)²Correct Option: A
As per the given in question , we draw a figure of two circles with radii 7 cm and 2 cm ,
As we know that , Direct common tangent = √(Distance between two centres)² - (r1 - r2)²
Direct common tangent = √(C1C2)² -(r1 - r2)²
Direct common tangent = √(13)² - (7 - 2)²
Direct common tangent = √169 - 25
Direct common tangent = √144 = 12 cm.
- The length of a tangent from an external point to a circle is 5√3 unit. If radius of the circle is 5 units, then the distance of the point from the circle is
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On the basis of question we draw a figure of a circle with centre O ,
Given that , AB = 5√3 units
OA = 5 units
∠ OAB = 90°
∴ OB = √AB² + OA²
OB = √(5√3)² + 5²Correct Option: A
On the basis of question we draw a figure of a circle with centre O ,
Given that , AB = 5√3 units
OA = 5 units
∠ OAB = 90°
∴ OB = √AB² + OA²
OB = √(5√3)² + 5²
OB = √75 + 25
OB = √100 = 10 units
∴ BC = OB – OC = 10 – 5 = 5 units
- PQ is a chord of length 8 cm, of a circle with centre O and of radius 5 cm. The tangents at P and Q intersect at a point T. The length of TP is
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As per the given in question , we draw a figure of a circle with centre O,
OT is the perpendicular bisector of chord PQ.
Let TR = y
∴ PR = QR = 4 cm
In right angle ∆ ORP,
OP² = OR² + PR²
⇒ OR² = OP² – PR² = 5² – 4² = 9
⇒ OR = 3 cm
In right angled ∆ PRT and ∆ OPT,
TP² = TR² + PR² and OT² = TP² + OP²
⇒ OT² = TR² + PR² + OP²
⇒ (y + 3)² = y² + 16 + 25⇒ 6y = 32 ⇒ y = 16 3
Correct Option: A
As per the given in question , we draw a figure of a circle with centre O,
OT is the perpendicular bisector of chord PQ.
Let TR = y
∴ PR = QR = 4 cm
In right angle ∆ ORP,
OP² = OR² + PR²
⇒ OR² = OP² – PR² = 5² – 4² = 9
⇒ OR = 3 cm
In right angled ∆ PRT and ∆ OPT,
TP² = TR² + PR² and OT² = TP² + OP²
⇒ OT² = TR² + PR² + OP²
⇒ (y + 3)² = y² + 16 + 25⇒ 6y = 32 ⇒ y = 16 3 ∴ TR = 16 3 ∴ TP² = TR² + PR² = 
16 
² + 16 3 TP² = 256 + 16 = 400 9 9 ∴ TP = 20 cm. 3
- The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and the bigger circle at E. Point A is joined to D. The length of AD is
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On the basis of question we draw a figure of two concentric circles with radii 13 cm and 8 cm ,
∠ODB = 90°
Here , OD = 8 cm and OB = 13 cm
∴ BD = √13² - 8²
BD = √169 - 64
BD = √105 cm
AE = 16 cm; ∠AED = 90°Correct Option: B
On the basis of question we draw a figure of two concentric circles with radii 13 cm and 8 cm ,
∠ODB = 90°
Here , OD = 8 cm and OB = 13 cm
∴ BD = √13² - 8²
BD = √169 - 64
BD = √105 cm
AE = 16 cm; ∠AED = 90°
AD = √AE² + DE²
AD = √256 + 105 = √361 = 19 cm
- From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then ∠APB is
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As per the given in question , we draw a figure of a circle in which two tangents PA and PB are drawn to with centre O ,
Let OA = OB = r and OP = 2r
⇒ AP = PBCorrect Option: D
As per the given in question , we draw a figure of a circle in which two tangents PA and PB are drawn to with centre O ,
Let OA = OB = r and OP = 2r
⇒ AP = PBsin ∠APO = OA = r = 1 OP 2r 2
∠ APO = 30°
∴ ∠ APB = 60°
