Chord in a Circle

let Q be any point on the arc AB opposite to point P . PA is tangent to the circle , by alternate segment theorem , $\angle PAB=\angle AQB$ . In right-angled triangle ABP , $\sin \angle PAB=\frac {PB}{AB}=\frac {6}{48}=\frac {1}{6}$ Hence , $\sin \angle AQB=\frac {1}{6}$ if the radius of the circle is R ; by sine law , $\frac {AB}{\sin \angle AQB }=2R$ $\Rightarrow R=\frac{48}{2.\frac{1}{6}}=3.48=144$ the radius is $144$ .

The angle between a tangent and the chord equals the angle subtended by that chord.

Assume angle BAP = x sinx = 1/6, then cosx = √35/6

The central angle will be 2x.

Let draw a perpendicular from the centre to the midpoint of AB, label it C, centre is D consider triangle ADC , DA is the radius.

AC = 24

AOC = x

sinx = 24/r

r = 144

Let the centre of the circle be $O$ . Firstly, let us find out some unknows from the given information. The only length we can find is $PA$ which, by applying the pythagoras' theorem turns out to be $8\sqrt{35}$ $$ Now, we know that $\\ \angle PBA = \angle PAO = 90^{\circ}$ . Therefore, $PBOA$ is a trapezium. $\\$ Now, drop a perpendicular $PS$ to $AB$ with $S$ on $AB$ . $\\$ Let $PS$ be $x$ . We know $AB=48 , \implies AS=SB=24$ . Therefore, $\boxed{\text{Radius}=OA=OB=\sqrt{x^2+24^2}}$ $$ Now, $[PBOA]=[PBA]+[BOA]$ Applying appropriate formulae, we get $\implies \frac{1}{2} \times (8+\sqrt{x^2+24^2})\times 8\sqrt{35} = (24x)+(\frac{1}{2}\times 8 \times 8\sqrt{35})$ $$ Solving for $x^2$ yields $x^2=35\times24$ $$ As already discussed earlier $\text{Radius}=\sqrt{(x^2+24^2)}$ . $$ Substituting the value of $x^2$ gives us $\boxed{\text{Radius}=144}$

Set angle BAP = X sin X = 8/48 = 1/6

Set central circle O Because segment line BP is parallel OA and draw a perpendicular from the centre to the midpoint of AB, label it C, so angle AOc=BAP consider triangle AOC , OA is the radius, r AC = 24 , and angle AOC = X

sin AOC = 24/r r = 24/sin P = 24/(1/6) = 144

construct the diameter AM, join M and B Let, the radius is r of right triangel ABM, MB^2=AM^2 - AB^2 \RightarrowMB= 2 \sqrt{r^2-24^2} PA^2 = AB^2 - PB^2 PA= 8 \sqrt{35} of right \DeltaBPA and \DeltaABM, \angle ABP =\angle MAB[MA and BP both are perpendicular on PA] so, \DeltaBPA and \DeltaABM are similar then, \frac {BP}{PA}=\frac {AB}{MB} \Rightarrow r= 144[by putting the values]

The angle PBA = arccos(1/6). Where A is the centre of the circle, the line OA is parallel to BP. This is because they are both at right angles to the line PA. BP by definition in the question and OA as the radius meets the tangent at a right angle. This all results in angle BAO being a right angle through alternate angles.

Using the cosine rule on triangle BOA:

r^2 = r^2 + 48^2 - 96rcos(arccos(1/6))

which leads to: r = 144.

Consider 'O' be the center of the circle, then AO is perpendicular to AP, extend chord PB so as to cut the circle at D once again and C is the mid point of chord BD. line segment OC is perpendicular to BD or PD. Now we have a rectangle OAPC

Applying Pythagoras Theorem, $AP^2$ = $AB^2$ - $PB^2$ = $48^2$ - $8^2$ = $2240$

Also using properties of tangents, $PA^2$ = $PB \times PD$

therefore,

PD= $PA^2$ / $PB$ = 2240/8= 280

hence, BD= PD-PB = 280-8= 272

Line OC bisects BD. therefore, BC= 272/2 = 136

So CP= BC+PB= 136+8= 144

since OAPC is a rectangle, OA = CP = 144 which is the distance of tangent from center which is indeed the radius. hence radius is 144. :-)

Let $C$ be a point on $\Gamma$ such that $AC$ is the diameter of $\Gamma$ . By Thales' theorem we have $\angle CBA = 90^\circ$ and since $PA$ is tangent to $\Gamma$ , thus $\angle CAP = 90^\circ$ . Therefore $\angle BAP = 90^\circ - \angle BAC = \angle ACB$ . Hence triangles $PAB$ and $BCA$ are similar by angle-angle-angle. So, we have $\frac{CA}{AB} = \frac{AB}{PB}$ $\Rightarrow CA = \frac{AB^2}{PB} = \frac{48 ^2}{8} = 288$ . This gives $r = \frac{288}{2} = 144$ .

You can solve it too easy..... Draw a perpendicular BC(C is a point on circle and on the opposite side of P) from AB.so AC is diameter. Remember that the triangles ABP & APC are similar.so BP/AB=AP/AC. Then we get that AC = 48*48/8.so the radii = 144.