Question 11: The figure shows two concentric circles and is the chord of larger circle. Prove that .

Answer:

If you drop a perpendicular from to the chord, it would bisect the chord *( Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.)*. Let us say that it intersects and at .

Therefore and

If you subtract these two, we get

Question 12: A straight line is drawn cutting two equal circles and passing through the midpoint of the line joining their centers . Prove that the chords , which are intercepted by two circles are equal.

Answer:

First draw perpendiculars and

Now consider and

(opposite angles)

is the mid point of – Given)

Therefore

Therefore

Therefore *(Theorem 8 (Converse of Theorem 7) : Chords of circle which are equidistant from the center of the circle are equal in length.)*

Question 13: are mid points of two equal chords respectively of a circle with center . Prove that:

(i)

(ii)

Answer:

First drop perpendiculars and on and respectively.

will bisect and will bisect . *( Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.)*

Therefore

Given . Therefore

Also

Therefore in (since angles opposite equal sides of a triangle are equal)

We know

(i) Therefore

(ii) We know

Therefore

Question 14: In the following figure; and are the points of intersection of two circles with centers and . If straight lines and are parallel to ; prove that

(i)

(ii)

Answer:

… … … … (i)

… … … … (ii)

Therefore from (i) and (ii) we get . Hence proved.

Question 15: Two equal chords and of a circle with center , intersect each other at point inside the circle. Prove that:

(i)

(ii)

Answer:

Draw and

*(Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.)*

In

In

(radius of the circle)

Consider

is common

Therefore

(Given)

Since

Therefore

Question 16: In the following figure, is a square. A circle is drawn with as the center which meets at and at . Prove that:

(i)

(ii)

Answer:

(i) In

(radius)

(side of square)

Therefore (S.A.S postulate)

(ii) (radius)

(side of a square)

Therefore

Consider

(side of a square)

Therefore

Question 17: The length of common chord of two intersecting circles is . If the diameter of the circles be and , calculate the distance between their center.

Answer:

and (given)

In

Similarly in

Therefore

Question 18: The line joining the mid points of two chords if a circle passes through the center. Prove that the chords are parallel.

Answer:

(given)

bisects (given)

Therefore (if line drawn from the center bisects the chord, then is is perpendicular to the chord)

Similarly

Therefore (alternate angles)

Therefore

Question 19: In the following figure, the line ; where and are the centers of the circles. Show that:

(i)

(ii)

Answer:

… … … … (i)

… … … … (ii)

(i) – (ii) we get

. Hence proved.

Question 20: and are two equal chords of a circle with center which intersect each other at right angle at point . If and , show that is a square.

Answer:

and (given)

Since and and because all angles are we can say

Therefore is square.

Question 21: In the given figure, is the center of the circle. and are two chords of circle. and . , . Find the

(i) radius of the circle

(ii) length of chord ** [2014]**

Answer:

, ,

Therefore

(i) Consider

Radius of the circle.

(ii) Now consider