If chords are drawn to the circle from a fixed point $(x _1,y _1)$ and then tangents are drawn at point of contact, the point of intersection of all tangents lie on a fixed point.

The radius perpendicular tangent at the pt. of contact, therefore, $OQ\perp PQ$ and $OR\perp PR$
In rt. $\triangle OPQ$, we have
$PQ=\sqrt{OP^{2}-OQ^{2}}$
   $=\sqrt{169-25}=\sqrt{144}=12$ cm
$\Rightarrow $ $PR=12$ cm (Two tangents from the same external pt. to a circle are equal)
Now area of quad. $PQOR=2\times $Area of $\triangle POQ$
   $\displaystyle =\left ( 2\times \frac{1}{2}\times 12\times 5 \right )$ cm$^{2}=60$ cm$^{2}$

The centres and radii of given circles are ${ C } _{ 1 }\left( 0,0 \right) ,{ r } _{ 1 }=4$ and ${ C } _{ 2 }\left( 0,1 \right) ,{ r } _{ 2 }=\sqrt { 0+1 } =1$
Now, ${ C } _{ 1 }{ C } _{ 2 }=\sqrt { 0+{ \left( 0-1 \right)  }^{ 2 } } =1$
and ${ r } _{ 1 }-{ r } _{ 2 }=4-1=3$
$\therefore { C } _{ 1 }{ C } _{ 2 }<{ r } _{ 1 }-{ r } _{ 2 }$
Hence, second circle lies inside the first circle.

Let length of tangent be $l$

Join O and P.
Now, in triangles OPQ and OPR,
OP = OP (Common)
OQ = OR (radius of circle)
PQ = PR (tangents from single point)

Hence OPQ and OPR are congruent triangles.
$\angle OPQ = \angle OPR = 30^{\circ}$

Since $\triangle ABO$ is congruent to $\triangle ACO$,  area of $ABOC$ is twice the area of $\triangle ABO$.

Since tangents and radii are perpendicular at the point of contact, in the quadrilateral formed by the two radii and the tangents at their ends, we have two right angles at the two points of contacts.

Two tangents can be drawn from an external point and both their lengths are equal.

Tangent is perpendicular to radius at the point of contact.

Circumference = 10 units

Let $P$ be $(h, k)$ then by $SS _{1} = T^{2}$ the equation of pair of tangents drawn from $P$ to $x^{2} + y^{2} = 1$ is
$(h^{2} + k^{2} - 1)(x^{2} + y^{2} - 1) = (hx + ky - 1)^{2}$
Its intersection with the line $x = 1$ is given by
$y^{2}(h^{2} + k^{2} - 1) = (ky + h - 1)^{2}$
or $y^{2}(h^{2} - 1) - 2yk (h - 1) - (h - 1)^{2} = 0 .....(1)$
It is a quadratic in $y$ and we are given that length of intercept is $1 \therefore y _{1} - y _{2} = 2$
or $(y _{1} + y _{2})^{2} - 4y _{1}y _{2} = 1$
or $\left [\dfrac {2k(h - 1)}{h^{2} - 1}\right ]^{2} + 4\dfrac {(h - 1)^{2}}{h^{2} - 1} = 4$
or $\dfrac {4k^{2}}{(h + 1)^{2}} + \dfrac {4(h - 1)}{h + 1} = 4$
or $k^{2} = (h + 1)^{2} - (h^{2} - 1) = 2h + 2$
Hence the locus of $(h, k)$ is $y^{2} =2(x + 1)$ which represents a parabola.

Given function,

$f(x)=1+c\left( x+3 \right)$                     .....( 1 )               where $c\in R$

We know that,

General equation of circle of radius $a$ meets at the origin is

${{x}^{2}}+{{y}^{2}}={{a}^{2}}$               

But it is unit circle then $a=1$,

Then, equation of circle is

$ {{x}^{2}}+{{y}^{2}}={{1}^{2}} $

$ {{x}^{2}}+{{y}^{2}}=1 $

Let  $f\left( x \right)=y$

By equation $\left( 1 \right)$ and we get,

$y=1+c\left( x+3 \right)$

$y=1+cx+3$                          ......( 2 )

Using distance formula,   at the origin to a line

$ d=\left| \dfrac{0-c\times 0-1-3c}{\sqrt{{{1}^{2}}+{{c}^{2}}}} \right|=1 $

$ \left| \dfrac{-1-3c}{\sqrt{{{1}^{2}}+{{c}^{2}}}} \right|=1 $

Taking square both side and we get, 

$ {{\left( \dfrac{1+3c}{\sqrt{1+{{c}^{2}}}} \right)}^{2}}=1 $

$ {{\left( 1+3c \right)}^{2}}=1+{{c}^{2}} $

$ {{1}^{2}}+{{\left( 3c \right)}^{2}}+6c=1+{{c}^{2}} $

$ 1+9{{c}^{2}}+6c=1+{{c}^{2}} $

$ 9{{c}^{2}}+6c-{{c}^{2}}=0 $

$ 8{{c}^{2}}+6c=0 $

$ 2c\left( 4c+3 \right)=0 $

$ 2c=0,4c+3=0 $

$ c=0,4c=-3 $

$ c=0,c=-\dfrac{3}{4} $

 Hence, It is required solution.

Since tangent is perpendicular to the radius through the point of contact
so, $PQ \bot OQ$

Given circle equation ${x}^{2}+{y}^{2}-2ax-2by+{b}^{2}=0$
Center $: (a,b)$ and Radius $= \sqrt { {a}^{2}+{b}^{2}-{b}^{2} } $
Both tangents are drawn from origin and perpendicular to each other. So, two tangent are $x$ and $y$ axis.
Hence, $\sqrt { {a}^{2}+{b}^{2}-{b}^{2} } = a = b$
$\Rightarrow {a}^{2}={b}^{2}$
$\Rightarrow {a}^{2}-{b}^{2} = 0$ 

$PA+PB \geq 2\sqrt {PA.PB}$   ...{A.M. $\geq $ G.M.}
$PA+PB \geq 2PT$   ...by tangent-secant theorem
$PA+PB \geq 2\sqrt{1+9-(-2)+12-8}=8$

let $(h,k)$ be the center
Then distance from center to both lines will be equal
$\left| \dfrac { h+2k }{ \sqrt { 5 }  }  \right| =\left| \dfrac { h-2k }{ \sqrt { 5 }  }  \right| $
$\Rightarrow hk=0$

Ans: A

Let $S\equiv { x }^{ 2 }+{ y }^{ 2 }-8x-6y+9=0$

The equation of pair of tangents drawn from the origin to the given circle are $S{ S } _{ 1 }={ T }^{ 2 }$
$\Rightarrow \left( { x }^{ 2 }+{ y }^{ 2 }-2px-2qy+{ g }^{ 2 } \right) \left( 0+0-0-0+{ g }^{ 2 } \right) ={ \left( x.0+y.0-p\left( x+0 \right) -q\left( y+0 \right) +{ y }^{ 2 } \right)  }^{ 2 }$
$\Rightarrow { q }^{ 2 }\left( { x }^{ 2 }+{ y }^{ 2 }-2px-2qy+{ g }^{ 2 } \right) -{ \left( -px-qy+{ g }^{ 2 } \right)  }^{ 2 }=0$
The two tangents are $\bot $ if ${ g }^{ 2 }+{ q }^{ 2 }-{ p }^{ 2 }-{ g }^{ 2 }=0$
(Sum of coefficient of ${ x }^{ 2 }+{ y }^{ 2 }=0$)
$\Rightarrow { q }^{ 2 }={ p }^{ 2 }$

The centers of the three given circles are $\left( -{ \alpha  } _{ 1 },0 \right) ,\left( -{ \alpha  } _{ 2 },0 \right) $ and $\left( -{ \alpha  } _{ 3 },0 \right) $.

The coordinates of $P$ be $(h,k)$. Then the equation of the tangents drawn from $P(h,k)$ to ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ is
$\left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) \left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 } \right) ={ \left( hx+hy-{ a }^{ 2 } \right)  }^{ 2 }$   (using SS'$={ T }^{ 2 }$)
This equation represents a pair of perpendicular lines.
Therefore, coefficient of ${ x }^{ 2 }$$+$ coefficient of ${ y }^{ 2 }=0$
$\Rightarrow \left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 }-{ h }^{ 2 } \right) +\left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 }-{ k }^{ 2 } \right) =0$
$\Rightarrow { h }^{ 2 }+{ k }^{ 2 }={ 2a }^{ 2 }$
Hence, locus is ${ x }^{ 2 }+{ y }^{ 2 }=2{ a }^{ 2 }$

Given equation of circle is $x^2+y^2-4x-2y-c=0$

From geometry we know $PA\cdot  PB = (PT)^{2}$

$t _{i}^{2} = x^{2} + y^{2} + 2g _{i}x + 5$  where $(x, y) $ is any point. 

Let $OA, OB$ be the tangents from the origin to the given circle with centre $C(-3, 5)$
and radius .$\sqrt{9 + 25 -c}= \sqrt{ 34 -c} $
Then area of the quadribiteral $ OACB = 2 \times $ area of  $\triangle OAC = 2 \times (\dfrac 12) \times OA\times AC $
Now $OA =$ length of the tangent from the origin to the given circle $ = .\sqrt{C}$
and $AC =$ radius of the circle $=\sqrt{ 34 -c} $ so that. $\sqrt{C} \sqrt{34 -c} =8 $       ...(given)
$\Rightarrow  c (34 -c) =64 \Rightarrow c^{2} -34c+64=0$

Equation of the given circle can be written as $(x -p)^{2} + (y -q)^{2} = p^{2}$
so, that the centre of the circle is $(p, q)$ and its radius is $p$.
This shows that $x = 0$ is a tangent to the circle from the origin.
Since tangents from the origin are perpendicular, the equation of the other tangent must be $y = 0$,
which is possible if $q = \pm  p $  or $p^{2} =q^{2}$

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.

Equation of the given circle can be written as ${ \left( x-r \right)  }^{ 2 }+{ \left( y-h \right)  }^{ 2 }={ p }^{ 2 }$
This has $(r,h)$ as the center and $r$ as the radius showing that it touches $y-$axis.
$\Rightarrow$ Other tangent from the origin to the circle must be $x-$axis which is possible if $h=\pm r$

The given circle is $S:{ x }^{ 2 }+{ y }^{ 2 }+x-2y-3=0$

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, $PA$ and $PB$ are the tangents from the point P.
$\angle APB = 54^{\circ}$
Now, In quadrilateral AOBP
$\angle OAP = \angle OBP = 90^{\circ}$ (Angle between tangent and radius)
Sum of angles = 360
$\angle OAP + \angle OBP + \angle OAB + \angle APB = 360$
$90 + 90 + 54 + \angle AOB = 360$
$\angle AOB = 126$

Now, In $\triangle OAB$
$OA = OB$ (Radius of circle)
$\angle OAB = \angle OBA$ (Isosceles triangle property)
Sum of angles = 180
$\angle OAB + \angle OBA + \angle AOB = 180$
$2 \angle OAB + 126 = 180$
$\angle OAB = 27^{\circ}$

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.

Let PA and PB are the tangents on the circle. $\angle APB = 60$. the radius of the circle with center at O be 3 cm.
The two tangents drawn to a circle from an external point are equally inclined to the segment joining the center to the point.
Thus, $\angle APO = 30^{\circ}$
In $\triangle OAP$
$\angle OAP = 90^{\circ}$       ...(Angle between tangent and radius)
$\tan 30 = \cfrac{1}{\sqrt{3}} = \dfrac{OA}{AP}$
$PA = 3 \sqrt{3}$

Let the coordinates of point$P$ be $\left( { x } _{ 1 },{ y } _{ 1 } \right). $

$5x-12y+10=0$ and $12y-5x+16=0$ are two parallel tangent to a circle.
Then distance $bet^{n}$ this two parallel tangents is $2r$.
$\therefore d=\left | \dfrac{-10-16}{\sqrt{5^{2}+12^{2}}} \right |=\left | \dfrac{26}{13} \right |=2$
$\therefore \ d=2r=2$
$\Rightarrow r=radius=1$

The equation of the tangent to the circle $x^2+y^2=4$ is

Equation of the circle is ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$    ...(1)

$(x-7)^2+(y+1)^2=25$
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}$

true 

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 

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

For two tangents to be drawn to $C(x, y):$

$\implies x^2 + y^2 +2x +2y – 16 = 0$

From P(k,k) , the point  P must lie outside the circle

$\implies C(k,k) > 0$

$\implies k^2 + k^2 + 2k + 2k – 16  > 0$

$\implies k^2 + 2k – 8 > 0$

$\implies (k + 4)(k - 2) > 0$

$\implies k \in (-\infty,-4) \cup (2,\infty)$

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

The equation of any line through the origin $(0,0)$ is

$y = mx +c$

If it is a tangent to the circle $(x−7)2+(y+1)2=52$, then

$ \dfrac{\left| 7m+1 \right|}{\left| \sqrt{{{m}^{2}}+1} \right|}=5 $

$ {{\left( 7m+1 \right)}^{2}}=5.\left( {{m}^{2}}+1 \right) $

$ 24{{m}^{2}}+14m-24=0 $

This equation, being a quadratic in m, gives two values of m, say ${{m} _{1}}$  and${{m} _{2}}$  These two values of m are the slopes of the tangents drawn from the origin to the given circle.

From (i), we have ${{m} _{1}}\times {{m} _{2}}=-1$

Hence, the two tangents are perpendicular.

Ans: C

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. $

Given:

Tangent is perpendicular to radius at the point of contact.

$x^2  + y^2 = (2\sqrt 2)^2 , C = (0,0) , r = 2 \sqrt 2$

Let $y = mx + 2\sqrt 2 \sqrt{m^2 + 1}$ be a tangent

To find the tangent through $(4,0)$ substitute into the equation

$\implies 0 = 4m +2\sqrt 2 \sqrt{m^2 + 1}$

$\implies 16m^2 = 8(m^2 + 1)$

$\implies m = -1$

Equation is

$y = -x + 4$

Substituting in circle equation

$x^2 + (-x + 4)^2 = 8$

$\implies 2x^2 – 8x + 8 = 0$

$\implies x = 2 \implies y = 2$

$A = (2,2)$

Any point on the circle is given be$ (2\sqrt2 \cos \theta, 2\sqrt2 \sin \theta )$

Let B = $(2\sqrt2 \cos \theta _1, 2\sqrt2 \sin \theta _1)$

$AB = 4 \implies AB^2 = 16$

$\implies (2\sqrt2 \cos \theta _1, -2)^2 + (2\sqrt2 \sin \theta _1 - 2)^2 = 16$

$\implies 8 + 4 + 4 – 4\sqrt 2(\cos \theta _1 + \sin \theta _1) = 16$

$\implies \sin \theta _1 + \cos \theta _1 = 0$

$\theta _1 = \dfrac{-\pi}{4}$

$B = (2\sqrt 2 \times\dfrac{1}{\sqrt 2} 2\sqrt 2 \times\dfrac{-1}{\sqrt 2})  = (2,-2)$

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

Diameter of circle=distance of the point (1,0)
from $3x-6y+7=0$
$\therefore$ $\cfrac { 3(1)-6(0)+7 }{ \sqrt { { \left( 3 \right)  }^{ 2 }+{ \left( -6 \right)  }^{ 2 } }  } =\cfrac { 10 }{ \sqrt { 45 }  } =\cfrac { 2 }{ 3 } \sqrt { 5 } $
Now, radius of circle $=\cfrac { 1 }{ 2 } \left( \cfrac { 2 }{ 3 } \sqrt { 5 }  \right) =\cfrac { \sqrt { 5 }  }{ 3 } $

• The radius of circle is $\sqrt{45} = 3\sqrt5$ , center of circle is $(0,0)$
• For the equation to be tangent to circle , the distance from center of circle to given line must be equal to radius of circle
• So we get $k/\sqrt5 = 3\sqrt5$ , which gives $k=15$

By intersecting secant theorem,

$x^2+y^2-2x=\dfrac {3}{5}\Rightarrow (x-1)^2+y^2=\dfrac {8}{5}$