Solution: Let’s draw a tangent \[TS\text{ }at\text{ }P\]to the circles given. As \[TPS\]is the tangent and \[PD\]is the chord, we have \[\angle PAB\text{ }=\angle BPS\text{ }\ldots .\text{ }\left( i...

### Two circles with centres O and O’ are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.

Solution: Join \[OA,\text{ }OB,\text{ }OA,\text{ }OB\text{ }and\text{ }OO\] \[CD\]is the tangent and \[AO\]is the chord. \[\angle OAC\text{ }=\angle OBA\text{ }\ldots \text{ }\left( i \right)\] ...

### Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.

Let \[DE\]be the tangent to the circle at \[P\] And, \[DE\text{ }||\text{ }QR\][Given] \[\angle EPR\text{ }=\angle PRQ\][Alternate angles are equal] \[\angle DPQ\text{ }=\angle PQR\][Alternate...

### AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.

Join \[OC\] \[\angle BCD\text{ }=\angle BAC\text{ }=\] \[{{30}^{o}}~\] [Angles in the alternate segment] It’s seen that, arc \[BC\]subtends \[\angle DOC\]at the center of the circle And \[\angle...

### If PQ is a tangent to the circle at R; calculate: i) ∠PRS ii) ∠ROT When O is the centre of the circle and ∠TRQ = 30degree

Solution: (i) As \[PQ\] is the tangent and \[OR\]is the radius. So, \[OR\bot PQ\] \[\angle ORT\text{ }=\text{ }{{90}^{o}}\] \[\angle TRQ\text{ }=\text{ }{{90}^{o}}-\text{ }{{30}^{o}}~=\text{...

### In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ADB = 30degree and ∠CBD = 60degree; calculate: ∠CDB

Solution: As \[BD\] is the diameter, we have \[\angle BCD\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle] Now in \[\vartriangle BCD\] \[\angle CDB\text{ }+\angle CBD\text{ }+\angle BCD\text{...

### In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ADB = 30degree and ∠CBD = 60degree; calculate: i) ∠QAB ii) ∠PAD

Solution: (i) Given, \[PAQ\]is a tangent and \[AB\]is the chord \[\angle QAB~=\angle ADB\text{ }=\text{ }{{30}^{o}}~\] [Angles in the alternate segment] (ii) \[OA\text{ }=\text{ }OD\][radii of the...

### In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find (i) AB. (ii) the length of tangent PT.

Solution: (i) \[PA\text{ }=\text{ }AB\text{ }+\text{ }BP\text{ }=\text{ }\left( AB\text{ }+\text{ }4 \right)\text{ }cm\] \[PC\text{ }=\text{ }PD\text{ }+\text{ }CD\] \[=\text{ }5\text{ }+\text{...

### In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

Solution: As \[PAB\]is the secant and \[PT\]is the tangent, we have \[P{{T}^{2}}~=\text{ }PA\text{ }x\text{ }PB\] \[{{12.5}^{2}}~=\text{ }10\text{ }x\text{ }PB\] \[PB\text{ }=\text{ }\left(...

### (i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP. (ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.

(i) (ii) Solution: (i) As the two chords \[AB\text{ }and\text{ }CD\] intersect each other at \[P\] We have \[AP\text{ }x\text{ }PB\text{ }=\text{ }CP\text{ }x\text{ }PD\] \[4.5\text{ }x\text{...