Circles — Exercise 10.2

The radius and tangent make a right angle. r²+24²=25², so r²=625−576=49.

Angles at P and Q in quadrilateral OPTQ are 90° each. Thus ∠PTQ=360°−90°−90°−110°.

RHS congruence of right triangles OAP and OBP shows OP bisects angle APB. Thus ∠OPA=40°. In right triangle OAP, ∠POA=90°−40°.

Let AB be a diameter, with centre O. The tangent at A is perpendicular to OA; the tangent at B is perpendicular to OB. OA and OB lie on the same straight line AB. Two lines perpendicular to the same line are parallel.

At contact point P, the radius OP is perpendicular to the tangent. Through a point on a line there is a unique perpendicular to that line. Therefore the constructed perpendicular coincides with OP and passes through O.

Join centre O to the contact point T and external point A. OT⊥AT, so OT²=OA²−AT²=25−16=9.

Let the chord be AB and contact point M. OM=3 and OA=5. OM⊥AB, so M bisects chord AB. AM=√(5²−3²)=4; AB=2AM.

Let contact points on AB,BC,CD,DA be P,Q,R,S. Tangents from a common external vertex are equal: AP=AS, BP=BQ, CQ=CR, DR=DS. Hence AB+CD=AP+BP+CR+DR=AS+BQ+CQ+DS=AD+BC.

From A, equal tangent lengths and RHS congruence show AO bisects the angle between the first parallel tangent and AB. Similarly BO bisects the interior angle at B between AB and the second parallel tangent. Those two interior angles sum to 180°, because the tangents are parallel. Their halves therefore sum to 90°. In triangle AOB, the remaining angle at O equals 180°−90°.

Let PA,PB be tangents and O the centre. ∠OAP=∠OBP=90°. The angle sum of quadrilateral OAPB is 360°, hence ∠APB+∠AOB=360°−180°.

For a tangential quadrilateral, AB+CD=AD+BC. In a parallelogram AB=CD and AD=BC. Substitute to get 2AB=2BC, hence adjacent sides are equal. Opposite-side equalities then make all four sides equal.

Let the two tangents from A each have length x. Equal tangents from B and C give AB=x+8, AC=x+6, BC=14. Semiperimeter s=x+14. The area equals rs=4(x+14). By Heron, area²=s(s−AB)(s−AC)(s−BC)=(x+14)×6×8×x. Equate: 16(x+14)²=48x(x+14). Since x+14>0, divide to get x+14=3x, hence x=7.

Let O be the centre and A,B,C,D the vertex angles. The centre lies on the internal angle bisector at each vertex, by RHS congruence of the contact-radius triangles. In triangle AOB, ∠AOB=180°−(A+B)/2. In triangle COD, ∠COD=180°−(C+D)/2. Since A+B+C+D=360°, their sum is 180°. The same calculation for BOC and DOA proves the other pair.