Chapter-10 Circles

Question No: 1

Create various circle pairs. How many points are shared by each pair? What is the most points that can be shared?

Explanation:


There is no shared point between these two circles.



Just point "P" is shared in this situation.


P is the common element even in this case.




P and Q are the two points that are shared here.


No two points in the circle above are similar.

Question No:2

Consider obtaining a circle. Permit the construction's center to be discovered.?

Explanation:



The following steps should be followed in order to locate the circle's centre:

Step I: First, make a circle.

Step ii: Draw the chords AB and CD in the circle.

step iii: Mark the perpendicular bisectors of AB and CD in step III.

Step iV: At a point, join the two perpendicular bisectors. The centre of the circle is where the two perpendicular bisectors intersect.

Question No:3

Three points at A, B, and C on a circle with centre O are shown in Fig. 10.36. BOC is at 30°, and AOB is at 60°. 

Find ADC if D is a location on the circle other than the ABC arc.




Explanation:

Considering that,

BOC added to AOB yields AOC.

AOC then equals 60° plus 30°.

∴ ∠AOC = 90°

The angle of an arc at its centre is well known to be twice as large as any other point on the circle.

So,

ADC = (1 + 2) AOC

= (½)× 90° = 45°

Question No:4

A circle's chord has a length equal to its radius. Both at a position on the minor arc and a point on the major arc, determine the angle the chord subtends.

Explanation:



In this case, the radius of the circle and the chord AB are equal. The circle's two radii are labelled OA and OB in the diagram above.

Now think about the OAB. Here,

AB = OA = OB = circle's radius

Hence, OAB is an equilateral triangle because it has all equal sides.

∴ ∠AOC = 60°

And ACB = 1/2 AOB.

Therefore, ∠ACB = ½ × 60° = 30°

A cyclic quadrilateral being ACBD,

The product of the angles ADB and ACB is 180 degrees.

Question No:5

Find BDC in Figure Given below where ABC = 69° and ACB = 31°.




Explanation:

If we are aware that all of the angles in the circle's section are equal,

BDC = BAC

Now. in the ΔABC, the sum of all the interior angles will be 180°.

ABC + BAC + ACB = 180 degrees.

Putting the values now

∠BAC = 180°-69°-31°

Therefore, ∠BAC = 80°

∴ ∠BDC = 80°

Question No:6

A, B, C, and D are four points on a circle in Fig. 10.39. At point E, AC and BD come together, with BEC = 130° and ECD = 20°. Locate BAC.



Explanation:

We are aware that the angles in the circle's section are equal.

So,

"BAC = CDE"

Using the triangle's feature of exterior angles,

In CDE, we receive

CEB equals CDE plus DCE

We are aware that 20° equals DCE.

Therefore, ∠ CDE = 110°

BAC and CDE are both equivalent.

∴ ∠ BAC = 110°.

Question No:7

Establish the existence of a rectangle if the diagonals of a cyclic quadrilateral are the diameters of the circle passing through its vertices.

Explanation:

Create a circle with the centre O and a cyclic quadrilateral ABCD whose diagonals AC and BD are both two diameters of the circle.


The semicircle's angles are known to be equal.

Hence, ABC = BCD = CDA = DAB = 90°

The quadrilateral ABCD is therefore a rectangle because each interior angle is 90.

Question No:8


Two places, B and C, where two circles cross. Two line segments ABD and PBQ are drawn via B to, respectively, intersect the circles at A, D, and P, Q. (see Fig. 10.40). demonstrate that ACP = QCD.



Explanation:

Construction:

AP and DQ should be joined.

We are aware that angles inside a segment are equal for chord AP.

Thus, PBA = ACP — (i)

The same goes for chord DQ.

A DBQ is a QCD (ii)

It is well known that the line segments ABD and PBQ cross at B.

The angles that are vertically opposed at B will be equal.

PBA = DBQ — (iii)

Combining equations I (ii), and (iii), we obtain

ACP equals QCD.

Question No:9

Prove that the point of intersection of these circles lies on the third side if circles are drawn using the two triangle sides as their diameters.

Explanation:

Draw a triangle ABC first, followed by two circles with diameters AB and AC, respectively.

Now we must show that D lies on BC and that BDC is a straight line.


Proof:

The semicircle's angles are known to be equal.

Hence, ADB = ADC = 90°.

Thus, ADB+ADC = 180°.

BDC follows a straight line.

As a result, D can be said to be on the line BC.

Using the two triangle sides as a guide, demonstrate that the point of intersection of these circles is on the third side.

Question No:10

Two right triangles with the same hypotenuse are ABC and ADC. Show that CBD = CAD.

Explanation:

We are aware that B = D = 90° and that AC is the common hypotenuse.

First, it must be demonstrated that CAD = CBD.


It can be claimed that ABC and ADC lie in a semi-circle because they are 90 degrees apart.

As a result, the semicircle contains the triangles ABC and ADC, and the points A, B, C, and D are concyclic.

Hence, CD is the circle's chord with O as its centre.

We are aware that the angles inside the same circle segment are equal.

CAD equals CBD.