(1)

$\angle OAP=60°$                          $[\because OA=OB=AB=\lambda ]$

$AP=\frac{2\lambda }{5} \text{(Given)}$

$\cos 60°=\frac{O{A}^2+A{P}^2-O{P}^2}{2·OA·AP}$

$\Rightarrow$$\frac{1}{2}=\frac{{\lambda }^2+\frac{4{\lambda }^2}{25}-O{P}^2}{2\lambda ·\frac{2\lambda }{5}}$

$\Rightarrow \frac{2{\lambda }^2}{5}=\frac{29{\lambda }^2}{25}-O{P}^2\Rightarrow O{P}^2=\frac{29}{25}{\lambda }^2-\frac{2}{5}{\lambda }^2=\frac{19}{25}{\lambda }^2$

Required locus is $x^2+y^2=\frac{19}{25}{\lambda }^2$

$\therefore$   $\text{Radius}=\frac{\sqrt{19}}{5}\lambda$

(2)

$\text{Since, }P(h,k)\text{ divides }AB\text{ in the ratio }3:2.$

So, $\frac{12\cos \theta +2}{5}=h \text{and} \frac{12\sin \theta +4}{5}=k$

$\Rightarrow 12\cos \theta =5h-2$                                        ...(i)

$\text{and }12\sin \theta =5k-4$                                      ...(ii)

Squaring (i) and (ii) and then adding, we get  

$144={(5h-2)}^2+{(5k-4)}^2$

$\Rightarrow {(x-\frac{{\displaystyle 2}}{{\displaystyle 5}})}^2+{(y-\frac{{\displaystyle 4}}{{\displaystyle 5}})}^2=\frac{144}{25}$

Centre $\equiv (\frac{{\displaystyle 2}}{{\displaystyle 5}},\frac{{\displaystyle 4}}{{\displaystyle 5}})\equiv C(\alpha ,\beta )$

$AC=\sqrt{{(1-\frac{{\displaystyle 2}}{{\displaystyle 5}})}^2+{(2-\frac{4}{5})}^2} =\sqrt{\frac{9}{25}+\frac{36}{25}}=\frac{3\sqrt{5}}{5}$

(3)

$\text{After solving }4x+3y=1\text{ and }3x-4y=32, \text{Point }A\text{ is }(4,-5)$

$\text{Centre }(\alpha ,\beta )\text{ lies on the line }4x+3y=1$

$\text{So, }4\alpha +3\beta =1$

$\therefore$  $\beta =\frac{1-4\alpha }{3}$

$\text{Now, distance from Centre to line }3x-4y-32=0$ $\text{and }3x+4y-24=0\text{ are equal.}$

$\text{So,}$$\left|\frac{{\displaystyle 3\alpha -4\left(\frac{{\displaystyle 1-4\alpha }}{{\displaystyle 3}}\right)-32}}{{\displaystyle \sqrt{3^2+4^2}}}\right|$$=$$\left|\frac{{\displaystyle 3\alpha +4\left(\frac{{\displaystyle 1-4\alpha }}{{\displaystyle 3}}\right)-24}}{{\displaystyle 5}}\right|$

$\Rightarrow$$\left|\frac{{\displaystyle 3\alpha -4\left(\frac{{\displaystyle 1-4\alpha }}{{\displaystyle 3}}\right)-32}}{{\displaystyle 5}}\right|$$=$$\left|\frac{{\displaystyle 3\alpha +4\left(\frac{{\displaystyle 1-4\alpha }}{{\displaystyle 3}}\right)-24}}{{\displaystyle 5}}\right|$

$\Rightarrow$$\left|\frac{{\displaystyle 9\alpha -4+16\alpha -96}}{{\displaystyle 15}}\right|$$=$$\left|\frac{9\alpha +4-16\alpha -72}{15}\right|$

$\Rightarrow 25\alpha -100=\pm (-7\alpha -68)$

$\therefore$ $\alpha =1 \text{and} \alpha =\frac{28}{3}$

For $\alpha =1$, Centre is $(1,-1)$

and $\text{radius is } \sqrt{{(4-1)}^2+{(-5+1)}^2}=5\text{ units}$

For $\alpha =\frac{28}{3}$, $\beta =\frac{-109}{9}, \text{Centre }$$(\frac{28}{3},\frac{-109}{9})$

$\therefore$ $\text{radius≃}8.88 \text{(rejected)}$

Hence, $\alpha =1, \beta =-1, r=5$

$\therefore$  $\alpha -\beta +r=7$

(2)

$C_1:x^2+y^2-18x-15y+131=0$

$C_2:x^2+y^2-6x-6y-7=0$

We know that for equation of circle  

$x^2+y^2+2gx+2fy+C=0$

Centre is $(-g,-f)$ and radius $=\sqrt{g^2+f^2-C}$

For $C_1,$ $\text{Centre }\left(A\right)=(9,\frac{{\displaystyle 15}}{{\displaystyle 2}})$

$\text{and radius }\left(r_1\right)=\sqrt{81+\frac{225}{4}-131}=\sqrt{\frac{25}{4}}=\frac{5}{2}$

For $C_2,$ $\text{Centre }\left(B\right)=(3,3)$

$\text{and radius }\left(r_2\right)=\sqrt{9+9+7}=5$

and $AB=\sqrt{{\left(6\right)}^2+{\left(\frac{9}{2}\right)}^2}=\sqrt{36+\frac{81}{4}}=\sqrt{\frac{225}{4}}=\frac{15}{2}=r_1+r_2$

$\therefore$ $\text{There are 3 common tangents}$

(1)

$\frac{dy}{dx}+\frac{x+a}{ y-2 }=0, y\left(1\right)=0;$    $\frac{dy}{dx}=\frac{-(x+a)}{ y-2 }$

$\Rightarrow \int (y-2)dy=-\int (x+a)dx\Rightarrow \frac{y^2}{2}-2y=\frac{-x^2}{2}-ax+c$

$\Rightarrow \frac{x^2+y^2}{2}+ax-2y=c\Rightarrow x^2+y^2+2ax-4y=c_1 (\because c_1=2c)$

$\text{Substituting }x=1\text{ and }y=0\text{ as }y\left(1\right)=0$

$\Rightarrow 1+2a=c_1 \therefore x^2+y^2+2ax-4y=1+2a$

$\text{This represents the equation of a circle having area }4\pi$

$\therefore$$\text{Radius of circle}=2$

$\text{Now }{r}^2=2^2={(-a)}^2+{\left(2\right)}^2=1+2a$

$\Rightarrow 0={(a+1)}^2 \Rightarrow a=-1$

$\therefore$ $\text{Equation of circle is } x^2+y^2-2x-4y=-1$

$\Rightarrow x^2-2x+1+y^2-4y+4-4=0$

$\Rightarrow {(x-1)}^2+{(y-2)}^2=4$

$\text{Now, substitute }x=0\text{ to get intersection point of circle with y-axis}$

$\Rightarrow {(y-2)}^2=3$

$\Rightarrow y=\pm \sqrt{3}+2$  $\therefore$$P(0,\pm \sqrt{3}+2)$

$P(0,\pm \sqrt{3}+2)\text{ and }Q(0, 2-\sqrt{3})\text{ are the points.}$

$\text{Now normal at }P\text{ will pass through centre }(1,2)$

$\therefore$ $\text{Equation of line passing through }P\text{ and }(1,2)\text{ is}$

$y-(\sqrt{3}+2)=\frac{-\sqrt{3}}{1}\left(x\right)$

$\text{Now, this line will cut x-axis at } R(\frac{{\displaystyle \sqrt{3}+2}}{{\displaystyle \sqrt{3}}}, 0)$

$\text{Also, equation of line passing through }Q(0, 2-\sqrt{3}) \text{ and }(1,2)\text{ is }$$y-(2-\sqrt{3})=\sqrt{3} x$

$\text{This will cut the x-axis at } S(\frac{{\displaystyle \sqrt{3}-2}}{{\displaystyle \sqrt{3}}}, 0)$

$RS=\sqrt{{(\frac{{\displaystyle \sqrt{3}+2}}{{\displaystyle \sqrt{3}}}-\frac{{\displaystyle \sqrt{3}-2}}{{\displaystyle \sqrt{3}}})}^2+{(0-0)}^2}=\sqrt{{\left(\frac{{\displaystyle 4}}{{\displaystyle \sqrt{3}}}\right)}^2}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$

 (1)

If a chord of circle of radius $R$ subtends angle ${\theta }_i$ at the centre $C_1,$ then locus of the end point of this chord in a circle of radius

$r_i=R\cos \left(\frac{{\displaystyle {\theta }_i}}{{\displaystyle 2}}\right)$

Given, $r_1^2=r_2^2+r_3^2$

$\Rightarrow {\cos }^2\frac{{\theta }_1}{2}={\cos }^2\frac{{\theta }_2}{2}+{\cos }^2\frac{{\theta }_3}{2}$

$\Rightarrow {\cos }^2\frac{\pi }{6}={\cos }^2\frac{{\theta }_2}{2}+{\cos }^2\frac{\pi }{3}$

$\Rightarrow \frac{3}{4}={\cos }^2\frac{{\theta }_2}{2}+\frac{1}{4}$

$\Rightarrow {\cos }^2\frac{{\theta }_2}{2}=\frac{1}{2}\Rightarrow \cos \frac{{\theta }_2}{2}=\frac{1}{\sqrt{2}}\Rightarrow \frac{{\theta }_2}{2}=\frac{\pi }{4}\Rightarrow {\theta }_2=\frac{\pi }{2}$

(2)

Clearly the line and circle intersect at (0, 0)  

$\Rightarrow \alpha =0$

$(1,\beta )$ lies on the circle  

$1+{\beta }^2-2\times 1=0 \Rightarrow {\beta }^2=1 \Rightarrow \beta =\pm 1$

Let $\beta =-1$

(1, -1) should lie on the line  

$\Rightarrow a-b=0\Rightarrow a=b\text{ but }a\ne b$

Therefore $\beta =1$.  

The line and circle intersect at (0, 0) and (1, 1).

Centre of circle with diameter AB is $(\frac{1}{2},\frac{1}{2})$

$\text{Radius of circle}=\frac{1}{\sqrt{2}}, \text{where }A(0,0)\text{ and }B(1,1)$

Let (p, q) be the mirror image of $(\frac{1}{2},\frac{1}{2})$ w.r.t. $x+y+2=0$

$\frac{p-1/2}{1}=\frac{q-1/2}{1}=\frac{{\displaystyle -2(\frac{{\displaystyle 1}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle 2}}+2)}}{{\displaystyle 2}}=-3$$\Rightarrow p=\frac{-5}{2}, q=\frac{-5}{2}$

(p, q) is centre of required circle whose equation is  

${(x+\frac{5}{2})}^2+{(y+\frac{{\displaystyle 5}}{{\displaystyle 2}})}^2=\frac{1}{2}\Rightarrow x^2+y^2+5x+5y+\frac{25}{2}-\frac{1}{2}=0$

$\Rightarrow x^2+y^2+5x+5y+12=0$

(2)

$\text{Equation of tangent at point }A(4,-11)\text{ is}$

$4x-11y-\frac{3}{2}(x+4)+5(y-11)-15=0$

$5x-12y=152$                                 ...(i)

$\text{Equation of tangent at point}B(8,-5)\text{ is}$

$8x-5y-\frac{3}{2}(x+8)+5(y-5)-15=0$

$\Rightarrow x=8$                                                ...(ii)

Solving (i) and (ii) we get

          $x=8 \text{and} y=\frac{-28}{3}$

So, point $C$ is $(8, -\frac{28}{3})$ and this point is centre of circle whose tangent is line $AB$.

Equation of line $AB$ is, $3x-2y=34$

Perpendicular distance of point $C$ from line $AB$ gives the radius of another circle whose centre is  $C(8,-\frac{{\displaystyle 28}}{{\displaystyle 3}}).$

So, $\frac{|3\times 8-2(\frac{-28}{3})-34|}{\sqrt{3^2+{(-2)}^2}}=\frac{2\sqrt{13}}{3}$

(1)

We have equation of circle ${(x-h)}^2+{(y-k)}^2=r^2$

We know that the perpendicular distance from centre to the tangent is equal to radius of the circle.

$L_1:x-y+2=0;$ $L_2:3x-4y+6=0;$ $L_3:4x-3y+1=0$

So, $r=\frac{h-k+2}{\sqrt{2}}=\frac{3h-4k+6}{\sqrt{9+16}}=\frac{4h-3k+1}{\sqrt{16+9}}$

Consider, $\frac{3h-4k+6}{5}=\frac{4h-3k+1}{5}$

$\Rightarrow 3h-4k+6=4h-3k+1\Rightarrow h+k=5$

(3)

We have, $x^2+y^2-4x-6y+11=0$                   ...(i)

Centre = (2, 3); Radius $=\sqrt{2}$

Tangent at $(3,2)$ is, $3x+2y-2(x+3)-3(y+2)+11=0$

$\Rightarrow x-y=1$

On rolling the given circle (i) upwards 4 units on the tangent $x-y-1=0$, centre of the circle also moves upwards 4 units on $T$. Let the centre of the new circle $C_1$ be $(h,k)$.

Since, slope of tangent = Slope of line joining two centres of the circle.  

Then the increment in both coordinates will be same.

From figure,

${(2+a-2)}^2+{(3+a-3)}^2=16$

$a^2+a^2=16 \Rightarrow a^2=8 \Rightarrow a=2\sqrt{2}$

Hence, the centre of circle $C_1$ is $(2+2\sqrt{2}, 3+2\sqrt{2})$ and radius remains same $\sqrt{2}$.

Equation of circle $C_1$ is ${(x-2-2\sqrt{2})}^2+{(y-3-2\sqrt{2})}^2=2$

The centre of circle $C_2$ is the image of $C_1$ in $T$, then 

$\frac{x-2-\sqrt{2}}{1}=\frac{y-3-2\sqrt{2}}{-1}=\frac{-2(2+2\sqrt{2}-3-2\sqrt{2}-1)}{1^2+{(-1)}^2}$

$\Rightarrow x-2-2\sqrt{2}=2 \text{and} y-3-2\sqrt{2}=-2$

$\Rightarrow y=1+2\sqrt{2} \Rightarrow x=4+2\sqrt{2}$

The centre of circle $C_2$ is  $B(4+2\sqrt{2}, 1+2\sqrt{2}).$

Feet of perpendicular from $A$ and $B$ on the $x$-axis are  $M(2+2\sqrt{2}, 0) \text{and} N(4+2\sqrt{2}, 0)$ respectively.

Area of trapezium AMNB $=\frac{1}{2}(3+2\sqrt{2}+1+2\sqrt{2})(4+2\sqrt{2}-2-2\sqrt{2})$

$=\frac{1}{2}\times (4+4\sqrt{2})\times 2=4+4\sqrt{2} \text{sq. units}.$