Peculiar Point P

This is the tangent chord theorem

$\Delta PTA \sim \Delta PBT$ , hence $PA / PT = PT / PB \Rightarrow PB = PT^2 / PA = 24^2/9=64$ .

According to the Power of a Point Theorem, PT^2 = PA * PB; 45^2 = 25 * PB; PB = 2025 / 25; PB = 81;

By the power of a point with respect to a tangent of a circle, we have that PT^2 = PA*(PB). So to solve for PB, we merely plug in and divide by PA. We get that the answer is 45^2/25= 81.

The Secant-Tangent Rule states that the (whole secant)×(external part) = (tangent's length)^2. So in this problem we have that the tangent's length is 45 and we have that the external part is 25, we are also looking for the whole secant (PB). Thus by plugging our values into the formula we get: (PB)(25)=(45)^2 This yield to PB= \frac {2025}{25} and thus out answer is PB=81

by tangent-secant theorem, (PT)(PT) = (PA)(PB) so, 45 * 45=25 * (PB) so, 2025 = 25*(PB) so, 81 = PB

hence, PB = 81

By Power of a Point, we know that PT^2 = PA*PB. Thus, PB = (PT^2)/(PA) =(45^2)/(25)=81.

By using tangent-secant theorem, which is the specific case of secant-secant theorem where there is a tangent line (refer to this ), we $PT^2 = PA \cdot PB$ .

Voila, we have $PB = \frac {PT^2}{PA}= \frac{45^2}{25} = 81$ .

using the power theorem on this case, then: the square of the distance from the external point to the point of tangency is equal to the product of the two divided parts of the secant

|PT|^2 = |PA| \times |PB|

(24)^2 = 9 \times |PB|

\frac {(24)^2}{9} = |PB|

64 = |PB|

line PB is a secant to the circle by property of similar triangles, PA.PB=PT.PT Let PB=x 9.x=24.24 x=64

According to a tangent property, the square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point.

=> PT^2 = PA.PB => 24^2 = 9.PB => PB = 64.

We know that $[pow _{\Gamma} (T)]^2=pow_{\Gamma} (A) \cdot pow_{\Gamma} (B)$ so $pow_{\Gamma} (B)=576/9=64$

PT^2 = PA PB so 45 45 = 25*PB so PB = 81

According to this formula, If a secant from P drawn to be meet the circle in two points A and B,then PA PB=PT PT. So, 25 PB=45 45 PB=9*9 PB=81

Let $S$ be the midpoint of $AB$ . The normal through $S$ meets the circle $Γ$ at the center $M$ .

Connect $M$ with $A$ to obtain a right angled triangle $\Delta ASM$ with $\angle ASM = 90 ^\circ$ and hypotenuse $MA$

$\Delta ASM$ introduces three unknowns: $AS=\frac{1}{2}AB \\ SM=h \\ MA=r$

Pythagorean theorem gives you:

$\left( \frac{1}{2}AB\right)^2+h^2=r^2 \tag{1}$

Consider the two right angled triangles $\Delta PSM$ and $\Delta PTM$ . They both share the same hypotenuse $PM$ .

For $\Delta PTM$ $PT=45 \\TM=r$

For $\Delta PSM$ $PS=\frac{1}{2}AB+25 \\ SM=h$

Pythagorean theorem for these two triangles gives: $(PS)^2+(SM)^2=(PM)^2=(PT)^2+(TM)^2$ or: $\left(\frac{1}{2}AB+25\right)^2+h^2=(45)^2+r^2 \tag{2}$

(1)-(2) leads to:

$\left(\frac{1}{2}AB\right)^2-\left(\frac{1}{2}AB+25\right)^2=-(45)^2$ Simplifying this expression leads to: $1400-25AB=0 \\ AB=56$

So the solution is: $PB=PA+AB \\ PB=25+56 \\ PB=81$

Just a direct application of Power of a Point. By Power of a Point in respect to P, we have $\overline{PT}^2=\overline{PA}\times\overline{PB}$ . Plugging in the values gives $45^2=25\times\overline{PB}$ so $\overline{PB}=\boxed{81}$

By power of $P$ respect to $\Gamma$ , $PA \cdot PB = PT^2$ , substituting the known values and solving gives us $PB = 81$

lets take the radius as x. A tangent makes a 90 degree angle with the center of the circle. The shortest length of PA is when PA produced becomes part of the diameter of the circle. Let's assume the center of the circle as O, then OA =OB=OT= the radius of the circle.

If we see the triangle PTO, ${PO}^2 = {PT}^2 + {OT}^2$ $\Rightarrow {25+x}^2 = {45}^2 + x^2$ $\Rightarrow 625 + x^2 +50x = 2025 + x^2$ $\Rightarrow 50x = 1400$ $\Rightarrow x =28$ .

Hence $PB = PA + AO + OB = 25 + 28 + 28 = 81$ .

By the Power Theorem, $(length of the tangent)^{2} = (length of the secant) \times (length of the secant outside the circle)$ .

Therefore $45^{2} = PB \times 25. PB = 81$

The following theorem has been used w.r.t(with respect to) the problem..........(PT)^2=(PA) x (PB) The theorem is called the power of a point . It goes like "One of the lines is tangent to the circle while the other is a secant (given figure). In this case, we have PB^2 = (PA) x (PB)." Plugging in 45 for PT and 25 for PA. 81 is got for PB.

There is a theorem that states that the square of the length of a tangent from an external point is equal to the product of the intercepts of the secant passing through the point.

Thus, PT^2=PA x PB, and 45^2=25*PB, PB=81.

By the power of a point, we have

$(PT)^2=PA(PB)$

$45^2=25(PB)$

$PB=\boxed{81}$

Let O be the centre of circle Γ.
Given that a line from P cuts Γ at A and B i.e. suppose that the line drawn from P also passes through the centre of the circle,O. Let the radius of circle Γ be r. Therefore, OT= r, OA =r (radius of the circle Γ) and OP is 9 + r (9 from the given fact that PA=9).

Since, Triangle TOP is a right-angled triangle (as the radius is perpendicular to the line tangent at any point on the circle).

Therefore using Pythagoras Theorem, OT^2 + PT^2 = OP*2; i.e. r^2 + 24^2 = (9+r)^2. Thus, r=27.5

Since, PB= r+r+PA i.e. PB = 27.5 + 27.5 + 9 = 64

PA.PB=PT^2 => PB=PT^2/PA=81

We have $PT^2 = PA.PB$. So $PB = \frac{PT^2}{PA} = 81$ So, my solution is 81

$PT^2 = PB \times PA$

$45^2 = PB \times 25$

$PB = 81$

by using the tangent law,

let PB = x ;

$45^{2}$ = (25)(x)

25x = 2025

x = 81

$45^{2}$ = $25 \times PB$

PB = 81

Using the theorem: The length of the tangent segment squared is equal to the product of the secant segment and its external segment. So 45^45= $25 \times x$

According to the power of a point, PA^2=PA*PB Therefore, PB=45x45/25=81