Circles

### Find the equation of a circle passing through the origin and intercepting lengths a and b on the axes.

Answer:           AD = b units and AE = a units. D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre. The general equation of a circle: (x - h)2 + (y - k)2 = r2...

### If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66o and ∠ABC = 80o. Calculate: (i) ∠DBC, (ii) ∠IBC,

Solution: Join $DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC$ Given, if $\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}$ $I$ is the incentre of the...

### In the given figure, ABC is a triangle in which ∠BAC = 30o. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.

Solution: Firstly, join $OB\text{ }and\text{ }OC$ Proof: $\angle BOC\text{ }=\text{ }2\angle BAC\text{ }=\text{ }2\text{ }x\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}$ Now, in $\vartriangle OBC$...

### In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of: (i) ∠AOB, (ii) ∠ACB,

Solution: (i) $Arc\text{ }AB$ subtends$\angle AOB$at the centre and $\angle ACB$at the remaining part of the circle. $\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB$...

### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: ∠PQR.

Join $OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS$ Given, $PQ\text{ }=\text{ }QR\text{ }=\text{ }RS$ So, $\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS$ [Equal chords subtends equal...

### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: (i) ∠POS, (ii) ∠QOR,

Solution: Join $OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS$ Given, $PQ\text{ }=\text{ }QR\text{ }=\text{ }RS$ So, $\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS$ [Equal chords subtends...

### Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.

Solution: Join $AC,\text{ }PQ\text{ }and\text{ }BD.$ As $ACQP$is a cyclic quadrilateral $\angle CAP\text{ }+\angle PQC\text{ }=\text{ }{{180}^{o}}~\ldots \ldots .\text{ }\left( i \right)$...

### Prove that: (i) the parallelogram, inscribed in a circle, is a rectangle. (ii) the rhombus, inscribed in a circle, is a square.

Solution: (i) Let’s expect that $ABCD$is a parallelogram which is inscribed in a circle. Hence, we know that $\angle BAD\text{ }=\angle BCD$  [Opposite angles of a parallelogram are equal] And...

### In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75Degree; ∠ABD = 58degree and ∠ADC = 77degree. Find: (i) ∠BDC, (ii) ∠BCD,

Solution: (i) According to the given question, By angle sum property of triangle ABD, $\angle ADB\text{ }=\text{ }{{180}^{o}}-\text{ }{{75}^{o}}-\text{ }{{58}^{o}}~=\text{ }{{47}^{o}}$ Hence,...

### Calculate: ∠ACB.

Solution: According to the given question By angle sum property of a triangle we have $\angle ACB\text{ }=\text{ }{{180}^{o}}-\text{ }{{49}^{o}}-\text{ }{{43}^{o}}~=\text{ }{{88}^{o}}$

### Calculate: (i) ∠CDB, (ii) ∠ABC,

Solution: According to the given question We get, (i) $\angle CDB\text{ }=\angle BAC\text{ }=\text{ }{{49}^{o}}$ (ii) $\angle ABC\text{ }=\angle ADC\text{ }=\text{ }{{43}^{o}}$ [Angles subtended...

### In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠OAB, (ii) ∠CBA.

Solution: According to the given question, (i) In $\vartriangle AOB,$we have $OA\text{ }=\text{ }OB\text{ }\left( radii \right)$ Thus, $\angle OBA\text{ }=\angle OAB$ By angle sum property of...

### In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

SOLUTION: (i) In the following figure, $\angle BAD\text{ }=\text{ }{{90}^{o}}~$ [Angle in a semi-circle] Therefore, $\angle BDA\text{ }=\text{ }{{90}^{o}}-\text{ }{{35}^{o}}~=\text{ }{{55}^{o}}$...

### In each of the following figures, O is the centre of the circle. Find the values of a, b and c.

(i) (ii) Solution: (i)In the following figure, $b\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{130}^{o}}$ [Angle at the center is double the angle at the circumference...

Solution: $\angle AOB\text{ }=\text{ }2\angle ACB$ [Angle at the center is double the angle at the circumference subtend by the same chord] $\angle ACB\text{ }=\text{ }{{70}^{o}}/\text{ }2\text{... read more ### In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45° (i) Prove that AC is a diameter of the circle. (ii) Find ∠ACB. Solution: (i) In \[\vartriangle ABD,$ $\angle DAB\text{ }+\angle ABD\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}$ ${{65}^{o}}~+\text{ }{{70}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}$...
Solution: Let’s join $AC.$ And, let $\angle OAC\text{ }=\angle OCA\text{ }=\text{ }x$ [Angles opposite to equal sides are equal] Thus, $\angle AOC\text{ }=\text{ }{{180}^{o}}-\text{ }2x$ Also,...