Geometry

Center is $(7,-1)$ and radius $=5$
Let equation of tangent from the origin be $y=mx$ $\Rightarrow mx-y=0$
Then, $\displaystyle\left| \frac { 7m+1 }{ \sqrt { { m }^{ 2 }+1 }  }  \right| =5$
$\Rightarrow { \left( 7m+1 \right)  }^{ 2 }=25\left( { m }^{ 2 }+1 \right) \Rightarrow 24{ m }^{ 2 }+14m-24=0$
Let ${ m } _{ 1 }$ and ${ m } _{ 2 }$ be the slopes of the two tangents.
Since $\displaystyle{ m } _{ 1 }{ m } _{ 2 }=-\frac{24}{24}=-1$
The two tangents are at right angles.

We know in $\triangle ABC, AB=5cm, BC=12cm$.
So, by pythagoras theorem we can find the length of side $AC$
$AC^2= AB^2 +BC^2=5^2+ 12^2$
$\therefore AC=13cm$
Circle is inscribed in a triangle. This type of circle is called as Incircle.
So, radius of incircle $=\displaystyle \frac {2 \triangle }{a+b+c}$
where $\triangle$ is the area of $\triangle ABC$ and $a,b,c$ are the sides of the triangle.
Area of $\triangle ABC= \displaystyle \frac {1}{2} AB \times BC= \frac {1}{2} \times 5 \times 12= 30sq.cm$
$\therefore$ radius of incircle $= \displaystyle \frac {2 \times 30}{5+12+13}=\frac {60}{30}=2cm$

Given, $BC = 6$ and $AB = 8$
using Pythagoras Theorem,
$AC^2 = AB^2 + BC^2$
$AC^2 = 6^2 + 8^2$
$AC = 10$
Radius = $\cfrac{2\times Area}{Perimeter}$
Radius = $\cfrac{2 \times (\dfrac{1}{2} \times 6 \times 8)}{10+8+6}$
Radius = $2$ cm

Let $y + 1 = m (x - 7) + \sqrt{25}(\sqrt{m^2 + 1})$ be any line to the circle.

Since we need tangents form $(0,0)$

$(0+1) = m(0 – 7) + 5\sqrt{m^2 + 1}$

$(7m + 1)^2 = 25(m^2 + 1)$

$\implies 24m^2 + 14m – 24 =0$

If $m _1, m _2$ are roots of the equation

$m _1m _2 = \dfrac{c}{a} = \dfrac{-24}{24} = -1$

Lines with $m _1$ and $m _2$ are slope are perpendicular.

Tangents from origin are at right angles to each other.

$(x-7)^2+(y+1)^2=25$
PA=PB=length of tangent from $(0,0) \space  to \space  (x-7)^2+(y+1)^2-25=0$
$=\sqrt{51}$
$\Rightarrow PA=PB=\sqrt{7^2+1-25}=5$
In $\Delta  OAP,$
$\tan  \alpha =\dfrac{OA}{PA}=\dfrac{5}{5}=1$
$\alpha =45^{\circ}$
So, angle both tangents $ =2\alpha =90^{\circ}$

By the power of a point theorem, for two secants PAB and PCD drawn from an external point P to a circle, the product of the segments of the secants is equal: PA * PB = PC * PD.

false, it is not always required it can even be less or greater than the radius of the circle, it depend on how far the point is from the center of the circle. 

true 

The angle between the two tangents and the angle between the radii at the points of contact are supplementary, as they form a quadrilateral with two 90-degree angles. Thus, 180 - 30 = 150 degrees.

Given $\angle BAC = 90^\circ$

$AB$ is tangent at $A$

$BDC$ is tangent at $D$

$CE = 3,CD =  6$

According to the tangent-secant theorem the length of tangent segment squared equals the product of secant segment and its external segment.

$\implies AC \times CE = CD^2$

$AC = \dfrac{6^2}{3} = 12$

$AE = 12 – CE = 9 = d = 2r$

Let $O$ be the center of the circle

$OA = OD = r$

$AB = BD = l$, tangents drawn from $B$

Since $\angle A = 90$

$BC^2 = AC^2 + AB^2$

$\implies (l + CD)^2 = AC^2 + l^2$

$\implies l^2 + 6^2 + 12l = 12^2 + l^2$

$\implies 12l = 108$

$BD = l = 9 \, cm = $ length of tangent

Area of triangle $ = \dfrac {RL^3}{L^2 + R^2}$
where R = radius of the circle = 3
L = length of tangent $ = \sqrt {S _1} = \sqrt {16 + 9 - 9} = 4$
Hence area $ = \dfrac {192}{25} sq.$ units

Let the angle between the tangents is $\theta\implies \tan \dfrac{\theta}{2}=\dfrac{r}{\sqrt{6^{2}+8^{2}-r^{2}}}$

$ Given-\ PA\quad &amp; \quad PB\quad are\quad tangents\quad to\quad the\quad circle\quad wtth\quad centre\quad O\ at\quad A\quad &amp; \quad B\quad respectively.\ To\quad find\quad out-\ The\quad assertion,\quad PA\quad or\quad PB\quad is\quad always\quad >\quad OA\quad or\quad OB,\quad is\quad \ true\quad or\quad false.\ Justification-\ PA\quad &amp; \quad PB\quad are\quad tangents\quad to\quad the\quad circle\quad at\quad A\quad &amp; \quad B\quad respectively.\ \therefore \quad PA\quad =\quad PB\quad and\quad \angle OAP\quad &amp; \quad \angle OBP\quad are={ 90 }^{ o }\ \therefore \quad \Delta OAP\quad is\quad a\quad right\quad one\quad with\quad \angle A={ 90 }^{ o }\ \Longrightarrow \angle AOP+\angle APO={ 90 }^{ o }\quad (by\quad angle\quad sum\quad prqperty\quad of\quad triangles)\ case\quad I-\quad \angle AOP=\angle APO\Longrightarrow each\quad of\quad them={ 45 }^{ o }.\quad i.e\quad PA=OA\ case\quad II-\quad \angle AOP>\angle APO\Longrightarrow \angle AOP>{ 45 }^{ o }\quad &amp; \quad \angle APO<{ 45 }^{ o }\quad \ (in\quad a\quad \Delta \quad the\quad side\quad opposite\quad to\quad the\quad greater\quad angle\quad is\quad greater\quad than\quad the\quad side\quad \ opposite\quad to\quad the\quad smaller\quad angle)\ \Longrightarrow PA>OA\ case\quad III-\quad \angle AOP<\angle APO\Longrightarrow \angle AOP{ <45 }^{ o }\quad &amp; \quad \angle APO>{ 45 }^{ o }\quad (\quad same\quad argument\quad as\quad case\quad II)\ \Longrightarrow PA>OA\ So\quad the\quad assertion,\quad PA\quad or\quad PB\quad is\quad always\quad <\quad OA\quad or\quad OB,\quad is\quad false.\ Ans-\quad False. $

$OT$ is a tangent and $OAB$ is a secant 

By intersecting secant theorem,