Theorems on circles - class-X

Two circles have two common tangents if the distance between their centers is less than the sum of their radii and greater than the difference of their radii. Here, centers are (0,0) and (0, lambda), radii are 2 and sqrt(lambda^2 - 5). The condition leads to the specified range.

For any two parallel tangents to a circle, the segment of a third tangent intercepted between them subtends a 90-degree angle at the center because the radii to the points of tangency are perpendicular to the tangents.

If angle NOQ = 120 degrees, then in the quadrilateral MONQ, the angles at N and Q are 90 degrees. Thus, angle M = 180 - 120 = 60 degrees. Triangle MNQ is isosceles with angle M = 60, so it is equilateral, meaning NQ = MN = MQ.

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

Tangents at the endpoints of a diameter are parallel to each other. Parallel lines intersect at an angle of 0 degrees.

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, $PQ \bot OQ$

The circumcircle of triangle TPQ has the segment OT as its diameter, where O is the center of the circle and T is the external point. If T lies on a line, the locus of the midpoint of OT will also be a straight line.

The power of a point (x1, y1) with respect to a circle x^2 + y^2 + 2gx + 2fy + c = 0 is x1^2 + y1^2 + 2gx1 + 2fy1 + c. For P(3,4) and S = x^2 + y^2 - 2y - 3 = 0, Power = 3^2 + 4^2 - 2(4) - 3 = 9 + 16 - 8 - 3 = 14.

This is a known identity in triangle geometry relating the lengths of tangents from the excenter to the circumcircle.

Using the condition for secants from a point to a circle and the given intercept length, the combined equation of the secants can be derived as x^2 - 4y^2 + 2x + 1 = 0.

A parallelogram that circumscribes a circle must have equal sums of opposite sides. Since opposite sides of a parallelogram are equal, this implies all four sides must be equal, which defines a rhombus.

The condition for a line y = mx + b to be tangent to a circle x^2 + y^2 - 2gx - 2fy + c = 0 is (mg + f + b)^2 = r^2(1 + m^2). For x^2 + y^2 - 6x - 16 = 0, the center is (3, 0) and r^2 = 16 + 9 = 25. Substituting g=3, f=0, r^2=25 gives (3m + b)^2 = 25(1 + m^2), not 5(1+m^2).

Given AB || CD, AB perpendicular to AD, and AB = 3CD, this is a right trapezoid. With area 4, we find the height and side lengths. A circle touches all sides if the sum of opposite sides is equal, which leads to the radius calculation via the geometry of the trapezoid.

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, $\sqrt { {a}^{2}+{b}^{2}-{b}^{2} } = a = b$
$\Rightarrow {a}^{2}={b}^{2}$
$\Rightarrow {a}^{2}-{b}^{2} = 0$ 

For a quadrilateral circumscribing a circle, the triangles formed by the center and the sides have properties such that the central angles subtended by opposite sides sum to 180 degrees.

The locus of points from which perpendicular tangents are drawn to a circle is the director circle. For x^2 + y^2 - 2x + 2y = 0, the center is (1, -1) and r^2 = 1 + 1 = 2. The director circle is (x-1)^2 + (y+1)^2 = 2(2) = 4. Point (2, sqrt(3)+1) gives (2-1)^2 + (sqrt(3)+1+1)^2 = 1 + (sqrt(3)+2)^2 = 1 + 3 + 4 + 4sqrt(3) = 8 + 4sqrt(3), which is not 4.

Let the radii of C1 and C2 be R and r, and their centers be O1 and O2. If a circle with center (x, y) and radius r' touches C1 internally and C2 externally, the distance to the centers relates to the radii, resulting in the definition of an ellipse.

The circle is (x+a)^2 + y^2 = r^2. A line y = mx through the origin is tangent to it. The distance from the center (-a, 0) to the line mx - y = 0 must equal r. Thus, |-ma| / sqrt(m^2 + 1) = r. Squaring gives m^2 a^2 = r^2(m^2 + 1), so m^2(a^2 - r^2) = r^2, leading to m = r / sqrt(a^2 - r^2).

$PA+PB \geq 2\sqrt {PA.PB}$   ...{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

Point P(1, sqrt(3)) lies on the circle x^2 + y^2 = 4 (radius 2). The secant PAB intersects the circle. The length PA + PB represents the sum of chord segments. The range is determined by the positions of the secant line.

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}$

The point (1, 2) satisfies x^2 + y^2 = 1 + 4 = 5. Since the point lies on the circle, exactly one tangent can be drawn at that point.

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.

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

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.

Tangents to the circle x^2 + y^2 = r^2 inclined at an angle of 45 degrees to the x-axis have a slope of m = tan(45 degrees) = 1 or -1. Using the formula y = mx + r*sqrt(1 + m^2), with r = 6, we get y = x plus or minus 6 times square root of 2.

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

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 

The circumcenter of triangle PQR, where P and Q are points of tangency from R, is the midpoint of the chord of contact and the center of the circle. Using the given circumcenter and radius, the equation can be derived.

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,

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