Circles - Exercise 10.2

Since radius is perpendicular to the tangent at the point of contact ($OP perp PQ$): $$angle OPQ = 90^circ$$ In right $Delta OPQ$: $$OQ^2 = OP^2 + PQ^2$$ $$13^2 = r^2 + 12^2$$ $$169 = r^2 + 144 implies r^2 = 169 - 144 = 25 implies r = sqrt{25} = 5 ext{ cm}$$

Since tangents are perpendicular to radii at points of contact: $$angle OPT = 90^circ quad ext{and} quad angle OQT = 90^circ$$ In quadrilateral $OPTQ$: $$angle OPT + angle POQ + angle OQT + angle PTQ = 360^circ$$ $$90^circ + 120^circ + 90^circ + angle PTQ = 360^circ$$ $$300^circ + angle PTQ = 360^circ implies angle PTQ = 360^circ - 300^circ = 60^circ$$

In $Delta OAP$ and $Delta OBP$: $OA = OB$ (radii), $PA = PB$ (tangents from $P$), and $OP = OP$ (common).
Therefore, $Delta OAP cong Delta OBP$ (SSS or RHS congruence).
Hence, $OP$ bisects $angle APB$: $$angle APO = rac{1}{2}angle APB = rac{1}{2}(60^circ) = 30^circ$$

In right-angled triangle $Delta OAP$ ($angle OAP = 90^circ$): $$angle POA + angle APO + angle OAP = 180^circ$$ $$angle POA + 30^circ + 90^circ = 180^circ implies angle POA = 180^circ - 120^circ = 60^circ$$

Let $AB$ be a diameter of a circle with centre $O$.
Let lines $PQ$ and $RS$ be the tangents drawn at endpoints $A$ and $B$ respectively.

Since radius through the point of contact is perpendicular to the tangent:
$OA perp PQ implies angle PAB = 90^circ$ and $angle QAB = 90^circ$.
$OB perp RS implies angle ABR = 90^circ$ and $angle ABS = 90^circ$.

Notice that $angle PAB = angle ABS = 90^circ$.
These are alternate interior angles made by the transversal line $AB$ with lines $PQ$ and $RS$.
Since alternate interior angles are equal, line $PQ parallel ext{line } RS$.

Let tangent $AB$ touch the circle with centre $O$ at point $P$.
By Theorem 10.1, $OP perp AB$, which means $angle OPB = 90^circ$.
Suppose, if possible, that the perpendicular to $AB$ at $P$ does not pass through centre $O$, but passes through another point $O'$.
Then, $angle O'PB = 90^circ$.
Comparing both: $angle OPB = angle O'PB = 90^circ$.
From the figure, this is possible only when ray $PO'$ coincides with ray $PO$.
Hence, the perpendicular at the point of contact must pass through the centre $O$.

Let $O$ be the centre and $T$ be the point of contact.
$OT$ is radius $r$, $AT = 8 ext{ cm}$, and $OA = 10 ext{ cm}$.
Since $OT perp AT$: $$OA^2 = OT^2 + AT^2$$ $$10^2 = r^2 + 8^2$$ $$100 = r^2 + 64 implies r^2 = 100 - 64 = 36 implies r = sqrt{36} = 6 ext{ cm}$$

Let chord $AB$ of the outer circle (radius $R = 13 ext{ cm}$) touch the inner circle (radius $r = 5 ext{ cm}$) at point $P$.
$OP perp AB$. Since the perpendicular from the centre to a chord bisects the chord: $$AP = PB implies AB = 2 imes AP$$

In right $Delta APO$: $$AO^2 = AP^2 + OP^2$$ $$13^2 = AP^2 + 5^2$$ $$169 = AP^2 + 25 implies AP^2 = 144 implies AP = 12 ext{ cm}$$

$$AB = 2 imes AP = 2 imes 12 = 24 ext{ cm}$$

Let the circle touch sides $AB, BC, CD$, and $DA$ at points $P, Q, R$, and $S$ respectively.
Since tangents drawn from an external point to a circle are equal in length: $$AP = AS quad ext{...(1)}$$ $$BP = BQ quad ext{...(2)}$$ $$CR = CQ quad ext{...(3)}$$ $$DR = DS quad ext{...(4)}$$

$$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$ $$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$ $$AB + CD = AD + BC$$

Let parallel tangents $XY$ and $X'Y'$ touch at $P$ and $Q$. Thus, $POQ$ is a straight line diameter.
Join $OC$.
In $Delta OPA$ and $Delta OCA$: $OP = OC$ (radii), $AP = AC$ (tangents from $A$), $OA = OA$ (common).
$Delta OPA cong Delta OCA implies angle POA = angle COA$.
Similarly, $Delta OQB cong Delta OCB implies angle QOB = angle COB$.

Since $POQ$ is a straight line: $$angle POA + angle COA + angle COB + angle QOB = 180^circ$$ $$2angle COA + 2angle COB = 180^circ$$ $$2(angle COA + angle COB) = 180^circ implies angle COA + angle COB = 90^circ$$ $$angle AOB = 90^circ$$

Let tangents $PA$ and $PB$ be drawn from external point $P$ to a circle with centre $O$. Points of contact are $A$ and $B$.
By Theorem 10.1: $OA perp PA implies angle OAP = 90^circ$ and $OB perp PB implies angle OBP = 90^circ$.
In quadrilateral $OAPB$: $$angle APB + angle AOB + angle OAP + angle OBP = 360^circ$$ $$angle APB + angle AOB + 90^circ + 90^circ = 360^circ$$ $$angle APB + angle AOB = 360^circ - 180^circ = 180^circ$$

Let $ABCD$ be a parallelogram circumscribing a circle.
From Question 8, for any circumscribed quadrilateral: $$AB + CD = AD + BC quad ext{...(i)}$$ Since $ABCD$ is a parallelogram, opposite sides are equal: $$AB = CD quad ext{and} quad AD = BC$$ Substituting these into (i): $$AB + AB = AD + AD$$ $$2AB = 2AD implies AB = AD$$ Since adjacent sides are equal in parallelogram $ABCD$, all four sides must be equal: $$AB = BC = CD = DA$$ Therefore, $ABCD$ is a rhombus.

Let points of contact on $AB$ and $AC$ be $E$ and $F$.
$BE = BD = 4 ext{ cm}$, $CF = CD = 3 ext{ cm}$.
Let $AE = AF = x ext{ cm}$.
Sides of $Delta ABC$: $a = BC = 4 + 3 = 7 ext{ cm}$, $b = AC = 3 + x$, $c = AB = 4 + x$.
Semi-perimeter $s = rac{a + b + c}{2} = rac{7 + (3+x) + (4+x)}{2} = rac{14 + 2x}{2} = 7 + x$.

Using inradius formula: $$ ext{Area} = r imes s = 2(7 + x)$$ Using Heron's Formula: $$ ext{Area} = sqrt{s(s-a)(s-b)(s-c)} = sqrt{(7+x)(x)(4)(3)} = sqrt{12x(7+x)}$$ Equating both expressions: $$sqrt{12x(7+x)} = 2(7+x)$$ Squaring both sides: $$12x(7+x) = 4(7+x)^2$$ Since $7+x eq 0$, divide both sides by $4(7+x)$: $$3x = 7 + x implies 2x = 7 implies x = 3.5 ext{ cm}$$

$$AB = 4 + x = 4 + 3.5 = 7.5 ext{ cm}$$ $$AC = 3 + x = 3 + 3.5 = 6.5 ext{ cm}$$

Let quadrilateral $ABCD$ circumscribe a circle with centre $O$, touching sides $AB, BC, CD, DA$ at $P, Q, R, S$ respectively.
Join $OP, OQ, OR, OS$, and vertices $OA, OB, OC, OD$.
$Delta OAP cong Delta OAS implies angle 1 = angle 8$.
$Delta OBP cong Delta OBQ implies angle 2 = angle 3$.
$Delta OCQ cong Delta OCR implies angle 4 = angle 5$.
$Delta ODR cong Delta ODS implies angle 6 = angle 7$.
Since sum of all central angles around $O$ is $360^circ$: $$(angle 1 + angle 8) + (angle 2 + angle 3) + (angle 4 + angle 5) + (angle 6 + angle 7) = 360^circ$$ $$2angle 1 + 2angle 2 + 2angle 5 + 2angle 6 = 360^circ$$ $$2[(angle 1 + angle 2) + (angle 5 + angle 6)] = 360^circ$$ $$(angle 1 + angle 2) + (angle 5 + angle 6) = 180^circ$$ $$angle AOB + angle COD = 180^circ$$ Similarly, $angle BOC + angle AOD = 180^circ$.