Advertisements
Advertisements
प्रश्न
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
Advertisements
उत्तर

Let O and O' be the centres of two intersecting circle, where
Points of intersection are P and Q and PA and PB are their diameter respectively.
Join PQ, AQ and QB.
∴ ∠AQP = 90° and ∠BQP = 90°
(Angle in a semicircle is a right angle)
Adding both these angles,
∠AQP + ∠BQP = 180°
∠AQB = 180°
Hence, the points A, Q and B are collinear.
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC

Prove that the parallelogram, inscribed in a circle, is a rectangle.
Prove that the rhombus, inscribed in a circle, is a square.
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠RNM

In the figure, given alongside, AB || CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
In the figure, given below, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of lengths 24 cm and 18 cm respectively.

In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC

In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.

In the given figure, AC is the diameter of the circle with center O.
CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°
Calculate:
- ∠BEC
- ∠BCD
- ∠CED

