Circle

The circle is $x^2+y^2=2^2$ (radius $r=2$ ).

• Intersection with $\mathbf{x}$-axis: Setting $y=0$ gives $x^2=4 \Rightarrow x= \pm 2$.

• Since $a>0, A=(2,0)$.

• Therefore, $B=(-2,0)$.

• Points $P$ and $Q$ :

• $P=(2 \cos \alpha, 2 \sin \alpha)$

• $Q=(2 \cos \beta, 2 \sin \beta)$

Let point of intersection of $A Q$ .& $B P$ be $R(h, k)$.

So, we need to find locus of points of intersection of $A Q \& B P$.

Slope of $B P=$ slope of $B R$

$ \begin{gathered} \frac{k-0}{h-(-2)}=\frac{2 \sin \alpha-0}{2 \cos \alpha+2} \\ \frac{k}{h+2}=\frac{\sin \alpha}{\cos \alpha+1} \Rightarrow \frac{k}{h+2}=\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^2 \frac{\alpha}{2}} \\ \frac{k}{h+2}=\tan \frac{\alpha}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{gathered} $

Similarly, slope of $Q A=$ slope of $Q R$

$ \begin{gathered} \frac{2 \sin \beta-0}{2 \cos \beta-2}=\frac{k-0}{h-2} \\ \frac{\sin \beta}{\cos \beta-1}=\frac{k}{h-2} \Rightarrow \frac{2 \sin \frac{\beta}{2} \cos \frac{\beta}{2}}{-2 \sin ^2 \frac{\beta}{2}}=\frac{k}{h-2} \\ -\cot \frac{\beta}{2}=\frac{k}{h-2} \\ \cot \frac{\beta}{2}=\frac{k}{2-h} \Rightarrow \tan \frac{\beta}{2}=\frac{2-h}{k} \end{gathered} $

Given,

$ \begin{gathered} \alpha-\beta=\frac{\pi}{2} \Rightarrow \frac{\alpha}{2}-\frac{\beta}{2}=\frac{\pi}{4} \\ \tan \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)=\tan \frac{\pi}{4} \\ \frac{\tan \frac{\alpha}{2}-\tan \frac{\beta}{2}}{1+\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}=1 \end{gathered} $

Substitute

$ \tan \frac{\alpha}{2}=\frac{k}{h+2}, \quad \tan \frac{\beta}{2}=\frac{2-h}{k} $

$ \begin{gathered} \frac{\frac{k}{h+2}-\frac{2-h}{k}}{1+\left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right)}=1 \\ \frac{k}{h+2}+\frac{h-2}{k}=1+\left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right) \\ \frac{k^2+h^2-4}{(h+2) k}=\frac{4 k}{(h+2) k} \\ \frac{k^2+h^2-4}{(h+2) k}=\frac{4 k}{(h+2) k} \\ k^2+h^2-4=4 k \\ k^2+h^2-4 k-4=0 \end{gathered} $

Replace $(h, k) \rightarrow(x, y)$

$ x^2+y^2-4 y-4=0 $

To find the value of $a-b$,

Determine the equation of circle $C_1$

The circle $C_1$ has a diameter of 10 , so its radius is $R=5$. It passes through the origin ( 0,0 ) and is contained in the closed half-plane $x \geq 0$.

Let the center of $C_1$ be $(h, k)$. Since it passes through $(0,0)$, we have $h^2+k^2=R^2=25$.

For the circle to lie entirely in the region $x \geq 0$ while passing through $(0,0)$, the distance from the center to the $y$-axis must be equal to the radius, and the center must have a non-negative $x$-coordinate.

Thus, $h=R=5$.

Substituting $h=5$ into $h^2+k^2=25$ gives $25+k^2=25$, which implies $k=0$.

The equation of $C_1$ is:

$ (x-5)^2+y^2=25 $

Find the center of circle $C_2$

The line $y=x$ is a chord of $C_1$. We find the intersection points of $y=x$ and $C_1$ :

$(x-5)^2+x^2 =25 $

$\Rightarrow $ $ x^2-10 x+25+x^2 =25 $

$\Rightarrow $$ 2 x^2-10 x =0 $

$\Rightarrow $ $ 2 x(x-5)=0 $

The intersection points are $(0,0)$ and $(5,5)$. These are the endpoints of the diameter of circle $C_2$.

The center $M$ of $C_2$ is the midpoint of the chord:

$ M=\left(\frac{0+5}{2}, \frac{0+5}{2}\right)=(2.5,2.5) $

Determine the equation of the chord of $C_2$

We are looking for the equation of a chord of $C_2$ passing through $P(2,3)$ that is farthest from the center $M(2.5,2.5)$.

In any circle, the chord passing through a point $P$ that is farthest from the center is the one perpendicular to the radius (or segment) $M P$.

The slope of the segment $M P$ is:

$ m_{M P}=\frac{3-2.5}{2-2.5}=\frac{0.5}{-0.5}=-1 $

The slope $m$ of the required chord is the negative reciprocal of $m_{M P}$ :

$ m=-\frac{1}{-1}=1 $

Using the point-slope form for the line through $P(2,3)$ with slope $m=1$ :

$ y-3 =1(x-2) $

$\Rightarrow $ $ y-3 =x-2 $ $\Rightarrow $ $ x-y+1 =0 $

Calculate $a-b$

The equation is given in the form $x+a y+b=0$. Comparing this to $x-y+1=0$, we identify:

$ \begin{aligned} & a=-1 \\ & b=1 \end{aligned} $

Finally, we calculate the required value:

$ a-b=-1-1=-2 $

Given a circle passes through $O(0,0), A(-\sqrt{3} a, 0)$, and $B(0,-\sqrt{2} b)$. Since $A$ lies on the x -axis and $B$ lies on the y -axis, $\angle A O B=90^{\circ}$. Thus $A B$ is the diameter of the circle.

Diameter $d=2 \times$ radius $=2(4)=8$.

Using distance formula for $A B$ :

$ \begin{gathered} \sqrt{(-\sqrt{3} a-0)^2+(0-(-\sqrt{2} b))^2}=8 \\ 3 a^2+2 b^2=64 \quad------(1) \end{gathered} $

Let the centroid of $\triangle O A B$ be $G(h, k)$.

$ \begin{gathered} h=\frac{0-\sqrt{3} a+0}{3} \Longrightarrow a=\frac{-3 h}{\sqrt{3}}=-\sqrt{3} h \\ k=\frac{0+0-\sqrt{2} b}{3} \Longrightarrow b=\frac{-3 k}{\sqrt{2}} \end{gathered} $

Substitute $a$ and $b$ into (Eq. 1):

$ \begin{gathered} 3(-\sqrt{3} h)^2+2\left(\frac{-3 k}{\sqrt{2}}\right)^2=64 \\ 3\left(3 h^2\right)+2\left(\frac{9 k^2}{2}\right)=64 \\ 9 h^2+9 k^2=64 \\ h^2+k^2=\frac{64}{9} \end{gathered} $

The locus is a circle $x^2+y^2=\left(\frac{8}{3}\right)^2$.

Radius $=\frac{8}{3} $

$ \begin{aligned} & \left|r_1-r_2\right| < c_1 c_2 < r_1+r_2 \\ & r-3 < \sqrt{9+25} < r+3 \\ & r \in(\sqrt{34}-3, \sqrt{34}+3) \\ & \alpha \beta=25 \end{aligned} $

$ \begin{aligned} & x^2+y^2-14 x-6 y-3=0 \\ & \tan 60^{\circ}=\frac{4}{\sqrt{h^2+k^2-4 h-6 k-3}} \end{aligned} $

squaring both side

$ 3\left(h^2+k^2-4 h-6 k-3\right)=16 $

To get locus replace $h, k$ by $x$ and $y$;

$ \begin{aligned} & 3 x^2+3 y^2-12 x-18 y-9-16=0 \\ & 3 x^2+3 y^2-12 x-18 y-25=0 \end{aligned} $

The equation of the parabola is $y = x^2.$

Differentiate to find the slope of the tangent: $\frac{dy}{dx} = 2x.$

Given that the slope is 4, $2x = 4 \Rightarrow x = 2.$

Substitute $x = 2$ into $y = x^2$: $y = (2)^2 = 4.$

Hence, $P(2, 4)$.

For the circle $x^2 + y^2 = 2$, differentiate: $2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}.$

Given that the slope is $-1$, $-\frac{x}{y} = -1 \Rightarrow x = y.$

Since $Q$ is in the first quadrant, substitute $x = y$ into the circle’s equation: $x^2 + x^2 = 2 \Rightarrow 2x^2 = 2 \Rightarrow x = 1, y = 1.$

Hence, $Q(1, 1)$.

For the ellipse $x^2 + 4y^2 = 8$, differentiate: $2x + 8y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{4y}.$

Given that the slope is $-\frac{1}{2}$, $-\frac{x}{4y} = -\frac{1}{2} \Rightarrow x = 2y.$

Since $R$ is in the first quadrant, substitute $x = 2y$ into the ellipse: $(2y)^2 + 4y^2 = 8 \Rightarrow 8y^2 = 8 \Rightarrow y = 1.$

Then $x = 2y = 2$, so $R(2, 1)$.

The triangle $ \triangle PQR $ is right-angled.

Now, find the length $PQ$: $PQ = \sqrt{(2 - 1)^2 + (4 - 1)^2} = \sqrt{10}.$

The radius of the circle passing through the three points is half the hypotenuse, so $\text{Radius} = \frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}.$

Since the circle touches the $x$-axis at a point which is $3$ units from the origin, the point of contact is $(3,0)$.

The centre lies in the first quadrant and the radius is perpendicular to the tangent at the point of contact. So, the centre must be of the form $(3,r)$, where $r$ is the radius.

The circle has an intercept of length $6\sqrt{3}$ on the $y$-axis. This means the chord cut by the line $x=0$ has length $6\sqrt{3}$.

For a circle of radius $r$, if the distance of a chord from the centre is $d$, then chord length is $2\sqrt{r^2-d^2}$.

Here, the distance from centre $(3,r)$ to the $y$-axis $(x=0)$ is $d=3$. So the intercept on the $y$-axis is

$2\sqrt{r^2-3^2}=6\sqrt{3}$

So,

$\sqrt{r^2-9}=3\sqrt{3}\ \Rightarrow\ r^2-9=27\ \Rightarrow\ r^2=36\ \Rightarrow\ r=6$

Hence, centre of the circle is $(3,6)$ and radius is $6$ units.

Now we need the length of the chord cut by the line $x-y=3$.

First find the perpendicular distance $p$ from the centre $(3,6)$ to the line $x-y-3=0$:

$p=\frac{|1\cdot 3+(-1)\cdot 6-3|}{\sqrt{1^2+(-1)^2}}=\frac{|3-6-3|}{\sqrt{2}}=\frac{6}{\sqrt{2}}$

Now chord length on this line is

$P Q=2 \sqrt{6^2-p^2} \quad \therefore p=\frac{6}{\sqrt{2}}$

$\begin{aligned} & \mathrm{p}=\frac{|3-6-3|}{\sqrt{2}} \\ & \mathrm{PQ}=2 \sqrt{36-\frac{36}{2}} \\ & =2 \times \frac{6}{\sqrt{2}}=6 \sqrt{2} \end{aligned}$

Vertex of the parabola

$ y=x^2-6x+12 $

is found by completing the square:

$ y=(x-3)^2+3 $

So, the vertex is

$ P=(3,3) $

Now the circle is

$ x^2+y^2-2x-4y+3=0 $

Complete squares:

$ (x-1)^2+(y-2)^2=2 $

Hence, the centre of the circle is

$ C=(1,2) $

and radius is

$ r=\sqrt{2} $

Now,

$ PC=\sqrt{(3-1)^2+(3-2)^2}=\sqrt{4+1}=\sqrt{5} $

Since the line through $P$ cuts the circle at $R$ and $S$, the line must intersect the circle in a chord $RS$.

Let the perpendicular distance from the centre $C$ to this line be $d$.

Then chord length

$ RS=2\sqrt{r^2-d^2}=2\sqrt{2-d^2} $

Also, since $P$ lies on the line outside the circle, and $R,S$ are the intersection points on opposite sides along the line, we have

$ PR+PS=RS $

only if $P$ lies between them, which is not possible here because $P$ is outside the circle.

So let us use power of a point.

If $PR$ and $PS$ are the two secant segments from $P$, then

$ PR \cdot PS = PC^2-r^2=5-2=3 $

Let the line make an angle such that the foot from $C$ to the line gives distance $d$.

If $M$ is the midpoint of chord $RS$, then

$ CM=d,\qquad PM=\sqrt{PC^2-d^2}=\sqrt{5-d^2} $

because in right triangle with hypotenuse $PC$.

Also

$ MR=MS=\sqrt{r^2-d^2}=\sqrt{2-d^2} $

Since $P$ is outside the circle, along the secant line,

$ PR=PM-MR,\qquad PS=PM+MR $

Therefore,

$ PR+PS = 2PM = 2\sqrt{5-d^2} $

So

$ (PR+PS)^2=4(5-d^2) $

To maximize this, we must minimize $d^2$.

The minimum possible value of $d$ is $0$, when the line passes through the centre $C$.

Hence maximum value is

$ (PR+PS)^2_{\max}=4\times 5=20 $

Therefore, the required maximum value is

$ \boxed{20} $

So the correct option is B.

First, find the vertex of the parabola.

The parabola is

$ x^2+6x+y+13=0 $

Rewrite it by completing the square:

$ x^2+6x+y+13=0 $

$ x^2+6x+9+y+13-9=0 $

$ (x+3)^2+y+4=0 $

$ y=-(x+3)^2-4 $

So, the vertex of the parabola is

$ (-3,-4) $

Now, find the centre and radius of the circle.

Given circle:

$ x^2+y^2-6x-8y+21=0 $

Complete the squares:

$ x^2-6x+y^2-8y+21=0 $

$ (x^2-6x+9)+(y^2-8y+16)+21-9-16=0 $

$ (x-3)^2+(y-4)^2-4=0 $

$ (x-3)^2+(y-4)^2=4 $

Hence, centre of the circle is

$ (3,4) $

and radius is

$ r=2 $

Now we need the maximum distance of a point $P$ on this circle from the vertex $(-3,-4)$.

Let $V=(-3,-4)$ and centre of circle be $C=(3,4)$.

First find $VC$:

$ VC=\sqrt{(3+3)^2+(4+4)^2} =\sqrt{6^2+8^2} =\sqrt{36+64} =\sqrt{100} =10 $

For a point moving on a circle, the maximum distance from an external point is

$ VC+r $

So,

$ \text{Maximum distance}=10+2=12 $

Therefore, the correct answer is

$ \boxed{12} $

So, Option C is correct.

The given circle is

$ x^2+y^2+x-3y=0 $

and the point from which the two chords are drawn is

$ P(1,2) $

We are told that these chords are bisected by the $y$-axis.

Let the other end of one such chord be $R(x,y)$.

Since the chord joining $P(1,2)$ and $R(x,y)$ is bisected by the $y$-axis, its midpoint must lie on the $y$-axis.

So the $x$-coordinate of the midpoint is $0$.

Midpoint of $P(1,2)$ and $R(x,y)$ is

$ \left(\frac{1+x}{2}, \frac{2+y}{2}\right) $

Since this lies on the $y$-axis,

$ \frac{1+x}{2}=0 $

So,

$ x=-1 $

Thus, the other end of each such chord must lie on the line

$ x=-1 $

Now find where this line cuts the circle.

Substitute $x=-1$ in

$ x^2+y^2+x-3y=0 $

We get

$ (-1)^2+y^2+(-1)-3y=0 $

$ 1+y^2-1-3y=0 $

$ y^2-3y=0 $

$ y(y-3)=0 $

So,

$ y=0 \quad \text{or} \quad y=3 $

Hence the two points are

$ R(-1,0), \quad S(-1,3) $

Now midpoint of $RS$ is

$ (\alpha,\beta)=\left(\frac{-1+(-1)}{2},\frac{0+3}{2}\right) $

$ (\alpha,\beta)=\left(-1,\frac{3}{2}\right) $

Therefore,

$ \alpha+\beta=-1+\frac{3}{2}=\frac{1}{2} $

So,

$ 6(\alpha+\beta)=6\cdot \frac{1}{2}=3 $

Hence the correct answer is

$ \boxed{3} $

So, the correct option is Option B.

$ \begin{aligned} & x+(k-1) y+3=0 \\ & 2 x+k^2 y-4=0 \\ & \left(\frac{1}{1-k}\right)\left(\frac{2}{k^2}\right)=1 \\ & 2=k^2-k^3 \\ & k^3-k^2+2=0 \\ & k=-1 \end{aligned} $

Solving :

$ \begin{aligned} & 2(x-2 y+3)=0 \\ & 2 x+y-4=0 \\ & -5 y+10=0 \\ & y=2 \end{aligned} $

$ x=1 $

Centre $(1,2)$

$ r=\sqrt{5} $

So circle is $(x-1)^2+(y-2)^2=5$

Chord $\mathrm{x}-\mathrm{y}+2=0$

$ \begin{aligned} & \mathrm{p}=\frac{1}{\sqrt{2}} \\ & \ell=\sqrt{5-\frac{1}{2}}=\frac{3}{\sqrt{2}} \\ & \mathrm{AB}=\frac{6}{\sqrt{2}} \quad \therefore \mathrm{AB}^2=18 \end{aligned} $

$ \begin{aligned} & \sqrt{16+36}=r+3 \\ & \Rightarrow r=\sqrt{52}-3 \\ & \Rightarrow \frac{3-3 r}{-r+3}=\alpha, \beta=\frac{9-3 r}{3+r} \\ & \Rightarrow \frac{(9-3 r-3+3 r)^2}{(r+3)^2} \\ & \Rightarrow \frac{36}{12}=\frac{9}{13}=\frac{m}{n} \Rightarrow m+n=22 \end{aligned}$

$\begin{aligned} 2 r & =\sqrt{52} \\ 4 r^2 & =52 \end{aligned}$

$\begin{aligned} &\begin{aligned} & 2 r^2=26 \\ & r^2=13 \end{aligned}\\ &\text { Eq }{ }^{\mathrm{n}} \text { of circle is }\\ &\begin{aligned} & x(x-4)+y(y-6)=0 \\ & k(k-4)+3 k(3 k-6)=0 \\ & k^2-4 k+9 k^2-18 k=0 \\ & 10 k^2=22 k \\ & 10 k=22 \\ & \therefore \quad 10 k+r^2=35 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \mathrm{M}_{A B}=0 \Rightarrow \mathrm{OM} \text { is vertical } \\ & \Rightarrow \alpha=2 \\ & \therefore \text { Centre }(0) \equiv(2,-4) \\ & \quad r=O A=\sqrt{(2-4)^2+(2+4)^2}=\sqrt{40} \end{aligned}\\ &\text { mid point of chord is } \mathrm{N} \equiv(1,2) \quad \therefore \mathrm{ON}=\sqrt{37}\\ &\begin{aligned} \therefore \text { length of chord } & =2 \sqrt{\mathrm{r}^2-(\mathrm{ON})^2} \\ & =2 \sqrt{40-37}=2 \sqrt{3} \end{aligned} \end{aligned}$

By solving $\mathrm{x}=\mathrm{y}$ with circle

We get

$\begin{aligned} & \mathrm{C}(\sqrt{2}, \sqrt{2}) \\ & \mathrm{D}(-\sqrt{2},-\sqrt{2}) \end{aligned}$

By solving $\mathrm{x}+\mathrm{y}=1$ with circle $x^2+y^2=4$ we set

$\mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right)$

& $\mathrm{B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right)$

$\therefore$ Area of Quadrilateral ACBD

$=2 \times$ Area of $\triangle \mathrm{BCD}$

$\begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{ccc} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1 \end{array}\right| \\ & =2 \sqrt{14} \end{aligned}$

By pytogorus $\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2$

$r=\sqrt{\frac{4 a^2+b^2}{4}}$

Equation of circle is $(x-\alpha)^2+(y-\beta)^2=r^2$

$\begin{aligned} & x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0 \\ & \text { comparision } x^2+y^2-\alpha x+\beta y+r=0 \\ & -\alpha=-2 a, \beta=-2 p, r=a^2 \\ & \Rightarrow 2 a=\alpha, 4 a^2+b^2=4 p^2 \\ & \alpha^2+b^2=4 p^2 \\ & \alpha^2+b^2=\beta^2 \end{aligned}$

So, $\left(2 a, b^2\right)=\left(\alpha, \beta^2-4 r\right)$

Centre $(1,-2), r=3$

Reflection of $(1,-2)$ about $2 x-3 y+5=0$

$\begin{aligned} & \frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2 \\ & x=-3, y=4 \end{aligned}$

Equation of circle ' $C$ '

$C:(x+3)^2+(y-4)^2=9$

A.T.Q.

$\begin{aligned} & \ell(\operatorname{arcAB})=\frac{1}{6} \times 2 \pi \mathrm{r} \\ & \mathrm{r} \theta=\frac{1}{6} \times 2 \pi \mathrm{r} \\ & \theta=\frac{\pi}{3} \\ & (\alpha+6)^2+(\beta-4)^2=27 \\ & \frac{(\alpha+3)^2 \pm(\beta-4)^2=9}{(\alpha+6)^2-(\alpha+3)^2=18} \\ & \Rightarrow 6 \alpha=-9 \\ & \Rightarrow \alpha=\frac{-3}{2}, \beta=\left(4-\frac{3 \sqrt{3}}{2}\right) \\ & \therefore \beta-\sqrt{3} \alpha \\ & \left(4-\frac{3 \sqrt{3}}{2}\right)+\frac{3 \sqrt{3}}{2} \\ & =4 \end{aligned}$

$ S_1:(x+2)^2+(y-2)^2=2^2 $

$ \mathrm{S}_2:(\mathrm{x}-2)^2+(\mathrm{y}-5)^2=\mathrm{r}^2 $

Both circle intersect at two points

$ \begin{aligned} & \therefore\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{c}_{\mathrm{c}} \mathrm{c}_2<\mathrm{r}_1+\mathrm{r}_2 \\ & |\mathrm{r}-2|<5<2+\mathrm{r} \\ & \Rightarrow 3<\mathrm{r}<7 \\ & \mathrm{r} \in(3,7) \\ & \alpha=3, \beta=7 \\ & 3 \beta-2 \alpha=15 \end{aligned} $

$ \begin{aligned} &\text { Let equation of circle be }\\ &\begin{aligned} & x^2+y^2+2 g x+2 f y=0 \\ & \Rightarrow \quad 0+\frac{100}{9}+2 f \times \frac{10}{3}=0 \\ & \Rightarrow \quad 2 f \times \frac{10}{3}=\frac{-10}{3} \times \frac{10}{3} \\ & \Rightarrow \quad f=\frac{-5}{3} \\ & \Rightarrow \quad 2^2+2^2+4 g+4 f=0 \\ & \Rightarrow \quad 8+4 g-\frac{20}{3}=0 \\ & \Rightarrow \quad 4 g=\frac{20}{3}-8=\frac{-4}{3} \Rightarrow g=-\frac{1}{3} \\ & \Rightarrow \quad r=\sqrt{g^2+f^2}=\sqrt{\frac{25}{9}+\frac{1}{9}}=\frac{\sqrt{26}}{3} \end{aligned} \end{aligned} $

$ \text { Let mid-point of chord } A B \text { be }(8, \alpha) $

Equation of chord $A B$ will be $T=S_1$

$ \begin{aligned} & 8 x+y \alpha-75=8^2+\alpha^2-75 \\ & 8 x+y \alpha=8^2+\alpha^2 \\ & \text { Slope }-\frac{-8}{\alpha} \text { and } 8^2+\alpha^2 \leq 75 \\ & \alpha^2 \leq 1 \end{aligned} $

Slope will be an integer for $\alpha= \pm 1, \pm 2$

Number of chord will be 4.

$ \text { Radius of circle } S=0 \text { is } C, P $

$ \begin{aligned} C_1 P & =\sqrt{5^2+2^2-19}=\sqrt{10} \\ C_1 & \equiv(5,2) \end{aligned} $

$ \begin{array}{ll} \Rightarrow & C_2 P=\frac{C_1 P}{2}=\frac{\sqrt{10}}{2} \\ \Rightarrow & C_1 C_2+C_2 P=C_1 P \\ \Rightarrow & C_1 C_2=\frac{C_1 P}{2} \\ \Rightarrow & C_1 C_2=C_2 P \end{array} $

Let $C_2 \equiv(\alpha, \beta)$

$C_2$ is the mid-point of $C_1 P$.

$ \alpha=\frac{5+2}{2}, \beta=\frac{2+3}{2} $

$ \begin{aligned} & \left(x-\frac{7}{2}\right)^2+\left(y-\frac{5}{2}\right)^2=\frac{10}{4} \\ & \Rightarrow(2 x-7)^2+(2 y-5)^2=10 \\ & \Rightarrow 4 x^2+4 y^2-28 x-20 y+49+25-10=0 \\ & \Rightarrow x^2+y^2-7 x-5 y+16=0 \end{aligned} $

$ \begin{aligned} &\text { } C A \cdot C P=r^2\\ &\begin{aligned} & \Rightarrow \sqrt{(2-1)^2+2^2} \times \sqrt{\left(2-\frac{7}{5}\right)^2+\left(0-\frac{6}{5}\right)^2}=r^2 \\ & \Rightarrow \sqrt{2^2+1^2} \times \sqrt{\frac{9}{25}+\frac{36}{25}}=r^2 \\ & \Rightarrow \sqrt{5} \times \frac{\sqrt{45}}{5}=r^2 \Rightarrow r^2=3 \Rightarrow r=\sqrt{3} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Let equation of circle } S=0 \text { be }\\ &\begin{aligned} & x^2+y^2+2 g x+2 f y+C=0 \\ & \Rightarrow \quad 2 g \times 2+2 f \times 0=c-7 \quad \rightarrow(\text { i }) \\ & \Rightarrow \quad 2 g \times 0+2 f \times \frac{1}{2}=c+0 \rightarrow(\text { ii }) \\ & \Rightarrow \quad 2 g \times \frac{3}{4} \times 2 f \times \frac{5}{4}=c-\frac{9}{2} \rightarrow(\text { iii }) \end{aligned} \end{aligned} $

$ \begin{array}{ll} & 4 g=c-7 \rightarrow(\mathrm{iv}) \\ & f=c \rightarrow(\mathrm{v}) \\ & 3 g+5 f=2 c-9 \rightarrow(\mathrm{vi}) \\ \Rightarrow & 3\left(\frac{c-7}{4}\right)+5 c=2 c-9 \\ \Rightarrow \quad & 3 c-21+20 c=8 c-36 \\ & 15 c=-15 \Rightarrow c=-1 \\ & g=-2, f=-1 \end{array} $

Equation of circle $s=0$

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

Equation of radical axis of $s=0$ and $s_1=0$ will be $s-s_1=0$.

$ \begin{array}{r} -8 x-2 y+6=0 \\ \Rightarrow \quad 4 x+y-3=0 \end{array} $

Equation of radical axis is

$ \begin{aligned} & s_1-s_2=0 \\ & (4 g-3) x+(4 f-8) y=0 \end{aligned} $

Centre of given circle $=(-1,-1)$

$ r=\sqrt{1^2+1^2-1}=0 $

Length of perpendicular from $(-1,-1)$ upon

$(4 g-3) x+(4 f-8) y$ equals radius of circle.

$ \begin{aligned} & \quad=\frac{|3-4 g+8-4 f|}{\sqrt{(4 g-3)^2+(8-4 f)^2}}=1 \\ & \Rightarrow 3-4 g+8-4 f \mid=\sqrt{(4 g-3)^2+(8-4)^2} \\ & \Rightarrow(3-4 g)(8-4 f)=0 \\ & \Rightarrow g=\frac{3}{4} \text { or } f=2 \end{aligned} $

Centre of circle is $O \equiv(1,-2)$

Radius of circle is $r=\sqrt{1^2+4+11}=4$

$ \begin{aligned} O D & =\sqrt{r^2-A D^2} \\ & =\sqrt{16-3}=\sqrt{13} \end{aligned} $

$O D=\sqrt{13}=$ Length of perpendicular from

$(1,-2)$ upon $2 x+3 y+k=0$

$ \begin{aligned} & \sqrt{13}=\frac{|2-6+k|}{\sqrt{13}} \\ & |k-4|=13 \\ & k-4= \pm 13 \\ & k=17 \text { and }-9 \end{aligned} $

∴ Sum of values of $k=17-9=8$

Let equation of circle be

$ x^2+y^2+2 g x+2 f y+c=0 $

and point is $(2,-1)$

$ \begin{array}{ll} \therefore & S_1=9 \\ \Rightarrow & 2^2+1^2+4 g-2 f+c=9 \\ \Rightarrow & 4 g-2 f+c=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } & -g-f=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \text { and } & g^2+f^2-c=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{array} $

$ \begin{array}{ll} \Rightarrow & g^2+f^2+4 g-2 f=20 \\ \Rightarrow & 2 g^2+4 g+2 g=20 \\ & 2 g^2+6 g=20 \\ \Rightarrow & g^2+3 g-10=0 \\ \Rightarrow & (g+5)(g-2)=9 \\ \Rightarrow & g=-5 \text { or } g=+2 \\ & -g=5 \text { or }-g=-2 \end{array} $

So, circle is $(-g,-f) \equiv(-2,2)$

So, $\beta-\alpha=4$

$ \begin{aligned} &\text { } 2 \theta=2 \tan ^{-1}\left(\frac{4}{3}\right)\\ &\tan \theta=\frac{4}{3} \end{aligned} $

$ \begin{aligned} & \sin \theta=\frac{4}{5}=\frac{r}{O P} \\ & \frac{4}{5}=\frac{\sqrt{9+9-2}}{O P} \\ \Rightarrow & O P^2=25 \\ \Rightarrow & (k+3)^2+(6 k-3)^2=25 \\ \Rightarrow & 37 k^2-30 k-7=0 \\ \Rightarrow & k=1 \end{aligned} $

$ \begin{aligned} C A & =r=\sqrt{2^2+1^2-1} \\ & =2 \end{aligned} $

$ \begin{aligned} C P & =\sqrt{4^2+2^2}=2 \sqrt{5} \\ \sin \theta & =\frac{2}{2 \sqrt{5}}=\frac{1}{\sqrt{5}} \\ \operatorname{Area}(\triangle A B C) & =\frac{1}{2} r^2 \sin (\pi-2 \theta) \\ & =\frac{1}{2} \times 4 \times 2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}} \\ & =\frac{8}{5} \end{aligned} $

$ \begin{aligned} \cos \theta & =\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2} \\ d & =C_1 C_2 \\ C_1 & \equiv(2,-1), C_2=(1,-2) \\ r_1 & =\sqrt{2^2+1^2+4} \\ r_2 & =\sqrt{1^2+2^2+11} \\ r_1 & =3, r_2=4 \\ \Rightarrow \quad d^2 & =1^2+1^2=2 \\ \Rightarrow \cos \theta & =\frac{9+16-2}{2 \times 3 \times 4}=\frac{23}{24} \\ \sin \theta & =\frac{\sqrt{47}}{24} \end{aligned} $

$ \begin{aligned} &\text { Equation of circle passing through }\\ &\begin{aligned} & x+y-2=0 \text { and } \\ & x^2+y^2+2 x-4 y+4=0 \text { is } \\ & x^2+y^2+2 x-4 y+4+\lambda(x+y-2=0 \\ & S=x^2+y^2+x(2+\lambda)+y(\lambda-4)+4-2 \lambda=0 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { The circle } S=0 \text { is orthogonal to }\\ &\begin{aligned} & x^2+y^2-2 x-4 y-4=0 \\ & \Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\ & \Rightarrow(2+\lambda)(-1)+(\lambda-4)(-2 \lambda=4-2 \lambda-4 \\ & \Rightarrow-2-\lambda-2 \lambda+8=-2 \lambda \\ & \Rightarrow \quad \lambda=6 \\ & \text { Radius of circle } S=0=\sqrt{4^2+1^2+8} \\ & \quad=5 \end{aligned} \end{aligned} $

Given, equation of circle

$ x^2+y^2+2 g x+2 f y-23=0 $

Centre $O(3,-2) \equiv(-g,-f)$

$ \begin{aligned} & g=-3, f=2 \\ & \text { Radius }=\sqrt{g^2+f^2-c}=\sqrt{9+4+23}=6 \\ & O P=\sqrt{(3+1)^2+(-2+5)^2}=\sqrt{16+9}=5 \\ & \because r>O P \end{aligned} $

∴ Minimum distance

$ \begin{aligned} & \frac{5 \alpha+3}{6}=-1 \Rightarrow \alpha=\frac{-9}{5} \\ & \frac{5 \beta-2}{6}=-5 \Rightarrow \beta=\frac{-28}{5} \\ \therefore & A\left(\frac{-9}{5}, \frac{-28}{5}\right) \end{aligned} $

Equation of circle

$ (x-a)^2+(y-a)^2=a^2 $

Passing through $A(1,2)$

$ (1-a)^2+(2-a)^2=a^2 $

$ \begin{aligned} &\begin{aligned} & a^2-6 a+5=0 \\ & a=1,5 \\ & S_1 \Rightarrow(x-1)^2+(y-1)^2=1 \\ & S_2 \Rightarrow(x-5)^2+(y-5)^2=5^2 \end{aligned}\\ &\text { Equation of } A B \Rightarrow S_1-S_2=0\\ &8 x+8 y-24-24=-24 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & x+y-3=0 \\ & L:\left(\frac{5 k+1}{k+1}, \frac{5 k+1}{k+1}\right) \end{aligned}\\ &\text { Lies on } A B \text {, }\\ &\begin{gathered} 10 k+2-3 k-3=0 \\ 7 k=1 \\ k=\frac{1}{7} \Rightarrow L=\left(\frac{3}{2}, \frac{3}{2}\right) \\ \therefore \quad A B=2 A L \\ =2 \sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{2}{\sqrt{2}}=\sqrt{2} \end{gathered} \end{aligned} $

Given, equation of circle

$ S: x^2+y^2-2 x+6 y+c=0 $

Centre $O(1,-3)$

$ \begin{aligned} r & =\sqrt{1+9-c}=\sqrt{10-c} \\ O P & =\left|\frac{4+9+2}{\sqrt{4^2+3^2}}\right|=3 \end{aligned} $

$ \begin{array}{ll} \therefore & O R=\sqrt{3^2+4^2}=5 \\ \Rightarrow & \sqrt{10-c}=5 \Rightarrow c=-15 \\ & S_1(1, k)=1^2+k^2-2+6 k-15=0 \\ \Rightarrow & k^2+6 k-16=0 \\ \Rightarrow & (k+8)(k+2)=0 \Rightarrow k=-8,2 \\ \because & k>0 \\ \therefore & k=2 \end{array} $

$(\text { Normal })_1=2 x-3 y+5=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$(\text { Normal })_2=4 x-5 y+7=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get

$ x=2, y=3 $

∵ Centre $O(2,3)$

∴ Radius $=\sqrt{(2-2)^2+(5-3)^2}=2$

Given equation of circles

$ \begin{aligned} & S_1: x^2+y^2+2 x-3 y-5=0 \\ & g_1=1, f_1=\frac{-3}{2}, c_1=-5 \\ & S_2: x^2+y^2-3 x+2 y+1=0 \\ & g_2=\frac{-3}{2}, f_2=1, c_2=1 \end{aligned} $

Let $S=x^2+y^2+2 \alpha x+2 \beta y+c=0$

$ \begin{aligned} \therefore \quad & 2 g g_1+2 f f_1=c_1+c_2 \\ & 2 \alpha-3 \beta=c-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 2 g g_2+2 f f_2=c_1+c_2 \\ & -3 \alpha+2 \beta=c+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ & (1,-1), 1+1+2 \alpha-2 \beta+c=0 \\ & 2 \alpha-2 \beta=-2-c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

By Eqs. (i) and (iii), By Eqs. (ii) and (iii),

$ \begin{array}{ll} -\beta=2 c-3 & -\alpha=-1 \\ \Rightarrow \beta=3-2 c & \Rightarrow \alpha=1 \end{array} $

By Eqs. (iii),

$ \begin{array}{ll} \Rightarrow & 2-2 \beta=-2-c \\ \Rightarrow & 2 \beta=c+4 \\ \Rightarrow & 2(3-2 c)=c+4 \\ \Rightarrow & 6-4 c=c+4 \\ \Rightarrow & 5 c=2 \Rightarrow c=\frac{2}{5} \\ \Rightarrow & \beta=3-\frac{4}{5}=\frac{11}{5} \\ & 5 \beta=11 \\ \therefore & \alpha-5 \beta=-10 \end{array} $

We have two circles given by:

$ \begin{aligned} & S_1: x^2+y^2-4 x-6 y-12=0 \\ & S_2: x^2+y^2+6 x+18 y+26=0 \end{aligned} $

We need to find a third circle that:

• Touches both $S_1$ and $S_2$ at their point of contact
• Passes through the point $(1, -1)$

Any circle that touches both $S_1$ and $S_2$ at their point of contact can be written as:

$ \left(x^2 + y^2 - 4x - 6y - 12\right) + \lambda \left(x^2 + y^2 + 6x + 18y + 26\right) = 0 $

Since the circle also passes through the point $(1, -1)$, substitute $x=1$, $y=-1$ into the equation: $ (1)^2 + (-1)^2 - 4(1) - 6(-1) - 12 + \lambda \Big[ (1)^2 + (-1)^2 + 6(1) + 18(-1) + 26 \Big] = 0 $

Calculate each term step-by-step:

• First bracket: $1 + 1 - 4 - (-6) - 12 = 2 - 4 + 6 - 12 = -8$ (Notice $-6 \times -1 = +6$)
• Second bracket: $1 + 1 + 6 - 18 + 26 = 2 + 6 - 18 + 26 = 8 - 18 + 26 = -10 + 26 = 16$

Solve for $\lambda$:

$ \lambda = \frac{8}{16} = \frac{1}{2} $

Now plug $\lambda = \frac{1}{2}$ into our general circle equation:

$ \begin{aligned} & \left(x^2 + y^2 - 4x - 6y - 12\right) + \frac{1}{2}\left(x^2 + y^2 + 6x + 18y + 26\right) = 0 \\ & x^2 + y^2 - 4x - 6y - 12 + \frac{1}{2}x^2 + \frac{1}{2}y^2 + 3x + 9y + 13 = 0 \\ & \left(1 + \frac{1}{2}\right)x^2 + \left(1 + \frac{1}{2}\right)y^2 + (-4 + 3)x + (-6 + 9)y + (-12 + 13) = 0 \\ & \frac{3}{2}x^2 + \frac{3}{2}y^2 - x + 3y + 1 = 0 \end{aligned} $

To make it simpler, multiply both sides by $2$:

$ 3x^2 + 3y^2 - 2x + 6y + 2 = 0 $

Divide all terms by $3$ to get standard form:

$ x^2 + y^2 - \frac{2}{3}x + 2y + \frac{2}{3} = 0 $

The center of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $(-g, -f)$.

From the equation:

• $2g = -\frac{2}{3}$ so $g = -\frac{1}{3}$
• $2f = 2$ so $f = 1$

$ \begin{aligned} & \Rightarrow(x-1)^2+(y-4)^2=\frac{(2 x+3 y-1)^2}{(4+9)} \\ & \Rightarrow\left(x^2+y^2-2 x-8 y+1+16\right)(13)=4 x^2 +9 y^2+1+12 x y-6 y-4 x \\ & \Rightarrow 9 x^2+4 y^2-12 x y-22 x-98 y+220=0 \end{aligned} $

$ \begin{aligned} & L_1: x+y=0 \\ & L_2: 2 x+y-1=0 \\ & L_3: x-3 y+2=0 \end{aligned} $

Now, equation of circumcircle

$ \begin{aligned} & \lambda_1 L_1 L_2+\lambda_2 L_2 L_3+\lambda_3 L_3 L_1=0 \\ & \lambda_1(x+y)(2 x+y-1)+\lambda_2(2 x+y-1) \\ & (x-3 y+2)+\lambda_3(x-3 y+2)(x+y)=0 \end{aligned} $

Coefficient of $x^2=$ Coefficient of $y^2$

$ \begin{aligned} & 2 \lambda_1+2 \lambda_2+\lambda_3=\lambda_1-3 \lambda_2-3 \lambda_3 \\ & \lambda_1+5 \lambda_2+4 \lambda_3=0 \\ & 5 \lambda_2=-\lambda_1-4 \lambda_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Coefficient of $x y=0$

$ \begin{aligned} & 3 \lambda_1-5 \lambda_2-2 \lambda_3=0 \\ & 5 \lambda_2=3 \lambda_1-2 \lambda_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i)and (ii),

$ \begin{aligned} \therefore \quad-\lambda_1-4 \lambda_3 & =3 \lambda_1-2 \lambda_3 \\ 4 \lambda_1 & =-2 \lambda_3 \Rightarrow \lambda_3=-2 \lambda_1 \\ \frac{\lambda_1}{\lambda_3} & =\frac{-1}{2} \end{aligned} $

And $5 \lambda_2=-\lambda_1+8 \lambda_1=7 \lambda_1$

$ \begin{array}{ll} & \lambda_2=\frac{7 \lambda_1}{5} \Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{5}{7} \\ \therefore & \frac{7 \lambda_1}{\lambda_2}+\frac{\lambda_3}{\lambda_1}=7 \times \frac{5}{7}-2=5-2=3 \end{array} $

$x_{\text {intercept }}=0$

$y_{\text {intercept }}=2$

Centre lies on $y=x+1$

Let equation of circle

$ x^2+y^2+2 g x+2 f y+c=0 $

Centre $(-g,-f) \Rightarrow-f=-g+1$

$ \begin{aligned} & f=g-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & x_{\text {intercept }}=2 \sqrt{g^2-c}=0 \\ & \Rightarrow \quad g^2=c \\ & y_{\text {intercept }}=2 \sqrt{f^2-c}=2 \\ & \Rightarrow \quad f^2-c=1 \\ & \\ & \\ & \\ & \quad f^2-g^2=1 \\ & \quad(f-g)(f+g)=1 \\ & \quad \quad f+g=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because f-g=-1]\\ & \quad\therefore f=-1, \quad g=0 \end{aligned} $

∴ Centre is $(0,1)$.

Which satisfies

$ x^2+y^2-26 x-20 y+19=0 $

Let equation of tangent be

$ y+2=m(x+1) \Rightarrow m x-y+m-2=0 $

∴ 1 st distance from centre $=$ radius

$ \begin{aligned} & \left|\frac{3 m-4+m-2}{\sqrt{1+m^2}}\right|=2 \\ & \Rightarrow(4 m-6)^2=4\left(1+m^2\right) \\ & \Rightarrow 3 m^2-12 m+8=0 \\ & \therefore \sqrt{3}\left|m_1-m_2\right|=\sqrt{3} \frac{\sqrt{D}}{a} \\ & =\sqrt{3} \frac{\sqrt{144-4 \times 8 \times 3}}{3} \\ & =\frac{4}{\sqrt{3}} \times \sqrt{3}=4 \end{aligned} $

$ \begin{array}{ll} \because & P A=P B \\ \therefore & \triangle O A P \cong \triangle O P B . \end{array} $

( $O$ is centre of circle

$ \begin{aligned} & x^2+y^2-4 x-4 y-8=0 \\ & r=\sqrt{4+4+8}=4) \end{aligned} $

∵ Equation of circle at $P$ and radius 4 is

$ (x-2)^2+(y+2)^2=2^2 $

$\therefore A B$ is common chord

∴ Equation is $S_1-S_2=0$

$ \begin{aligned} &\left(x^2+y^2-4 x-4 y-8\right) \\ & -\left(x^2+y^2-4 x+4 y+4+4-4\right)=0 \\ & \Rightarrow \quad-8 y-8-4=0 \\ & \Rightarrow \quad 8 y+12=0 \\ & 2 y+3=0 \end{aligned} $

∵ If two circles cut orthogonally.

$ \begin{aligned} & \Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1 c_2 \\ & S_1: x^2+y^2-2 y-3=0 \\ & S_2: x^2+y^2+4 x+3=0 \\ & S_3: x^2+y^2-2 \alpha x-2 \beta x+c=0 \rightarrow \text { centre } \\ & (\alpha, \beta) \\ & \therefore 2(-\alpha)(0)+2(-\beta)(-1)=c-3 \\ & \quad 2 \beta=c-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

And $2(-\alpha)(2)+2 \beta(0)=c+3$

$ -4 \alpha=c+3 \Rightarrow 4 \alpha=-c-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 4 \alpha+2 \beta=-6 \\ \Rightarrow \quad & 2 \alpha+\beta=-3 \end{aligned} $

Circle $C_1 \rightarrow$ radius $r_1, O_1(1,2), r_1=3 r^{\prime}$

Circle $\mathrm{C}_2 \rightarrow$ radius $r_2, \mathrm{O}_2(3,-2), r_2=r^{\prime}$

$\because C_1$ and $C_2$ cuts orthogonally

$ \begin{aligned} & r_1^2+r_2^2=0_1 0_2^2 \\ \Rightarrow \quad & \left(3 r^{\prime}\right)^2+r^{\prime 2}=2^2+4^2 \\ \Rightarrow \quad & 10 r^{\prime 2}=20 \Rightarrow r^{\prime}=2 \end{aligned} $

∴ Equation of required circle

$ \begin{aligned} & \Rightarrow(x-1)^2+(y+2)^2=2 \\ & \Rightarrow x^2+y^2-2 x+4 y+3=0 \end{aligned} $

Given, equation of circles

$ x^2+y^2=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

And $(x-9)^2+y^2=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} C_1 & =(0,0), r_1=4 \\ C_2 & =(9,0), r_2=4 \\ \therefore \quad C_1 C_2 & =\sqrt{(9-0)^2+(0-0)^2} \\ & =\sqrt{(9)^2}=9 \end{aligned} $

And $r_1+r_2=4+4=8$

Since $C_1 C_2>r_1+r_2$

Circles are separate to each other, any tangent to circle (i) with slope ( $m$ ) is

$ \begin{aligned} & y=m x \pm r \sqrt{1+m^2} \\ \Rightarrow \quad & y=m x+4 \sqrt{1+m^2} \\ \Rightarrow \quad & m x-y \pm 4 \sqrt{1+m^2}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Equation (iii) to be tangent to circle (ii)

$ \begin{array}{ll} \therefore & \left|\frac{m(9)-0 \pm 4 \sqrt{1+m^2}}{\sqrt{m^2+1}}\right|=4 \\ \Rightarrow & 16\left(m^2+1\right)=\left(9 m \pm 4 \sqrt{1+m^2}\right)^2 \\ \Rightarrow & 16 m^2+16=81 m^2+16\left(1+m^2\right) \\ & \pm 72 m \sqrt{1+m^2} \end{array} $

$ \begin{array}{ll} \Rightarrow & \quad 81 m^2 \pm 72 m \sqrt{1+m^2}=0 \\ \Rightarrow & 9 m\left(9 m \pm 8 \sqrt{1+m^2}\right)=0 \end{array} $

When $9 m=0 \Rightarrow m=0$

Then Eq. (iii) $y= \pm 4$ (horizontal lines)

If, $9 m \pm 8 \sqrt{1+m^2}=0$

$ \begin{aligned} & \Rightarrow \quad(9 m)^2=\left( \pm 8 \sqrt{1+m^2}\right)^2 \\ & \Rightarrow \quad 81 m^2=64\left(1+m^2\right) \\ & \Rightarrow \quad 81 m^2-64 m^2=64 \\ & \Rightarrow \quad 17 m^2=64 \\ & \Rightarrow \quad m= \pm \sqrt{\frac{64}{17}} \\ & \Rightarrow \quad m= \pm \frac{8}{\sqrt{17}} \end{aligned} $

$ \begin{aligned} &\text { Given, equation of circle }\\ &\begin{aligned} & x^2+y^2-4 x-6 y-12=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { centre }(-g,-f)=(2,3) \\ & \text { And radius }=\sqrt{g^2+f^2-c} \\ & \qquad \begin{aligned} & =\sqrt{4+9+12}=\sqrt{25}=5 \end{aligned} \end{aligned} \end{aligned} $

Equation of new circle whose radius is 3

$ \begin{aligned} & (x-h)^2+(y-k)^2=(3)^2 \\ & \Rightarrow \quad(x-h)^2+(y-k)^2=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Since circle (i) and (ii) touches each other internally at $(-1,-1)$

$ \therefore \quad C_1 C_2=\left|r_1-r_2\right| $

where $c_2=(h, k)$

$ \therefore \quad C_1 C_2=|5-3|=2 $

$P$ divides the line segment $C_1 C_2$ in the ratio $r_1: r_2$

$ \quad \begin{aligned}\therefore C_1 P & =r_1 \text { and } C_2 P=r_2 \\ C_1 P & =\sqrt{(-1-2)^2+(-1-3)^2} \\ & =\sqrt{(-3)^2+(-4)^2} \\ & =\sqrt{9+16}=\sqrt{25}=5 \end{aligned} $

and $C_2 P=r_2=3$

$\therefore$ The point of tangency $P$ divides the line segment $C_1 C_2$ externally in the ratio $r_1:-r$

$ \begin{array}{rlrl} \therefore & P =\frac{r_1 C_2-r C_1}{r_1-r} \\ & =\frac{5 C_2-3 C_1}{5-3}=\frac{5 C_2-3 C_1}{2} \\ \Rightarrow & 2 P =5 C_2-3 C_1 \\ \Rightarrow & 5 C_2 =2 P+3 C_1 \\ \Rightarrow & 5(h, k) =2(-1,-1)+3(2,3) \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 5(h, k)=(-2,-2)+(6,9) \\ & \Rightarrow \quad 5(h, k)=(4,7) \Rightarrow(h, k)=\left(\frac{4}{5}, \frac{7}{5}\right) \end{aligned}\\ &\therefore \text { Equation of required circle }\\ &\begin{aligned} & \left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=(3)^2 \\ & \Rightarrow \quad\left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=9 \\ & \Rightarrow \quad x^2+\frac{16}{25}-\frac{8}{5} x+y^2+\frac{49}{25}-\frac{14}{5} y=9 \\ & \quad x^2+y^2-\frac{8}{5} x-\frac{14}{5} y+\frac{13}{5}=9 \\ & \Rightarrow \quad 5 x^2+5 y^2-8 x-14 y-32=0 \end{aligned} \end{aligned} $

If $C_1$ and $C_2$ are two circles having no common points, it means they do not intersect and do not touch each other then there will be no common tangents or there will be four common tangents $C_1$ and $C_2$

Equation of the circle touching the $X$-axis and passing through the point $(-1,1)$

$ \begin{array}{ll} & (-1-h)^2+(1-k)^2=k^2 \\ \Rightarrow & 1+h^2+2 h+1+k^2-2 k=k^2 \\ \Rightarrow & 1+2 h+h^2+1-2 k+k^2=k^2 \\ \Rightarrow & 1+2 h+h^2+1-2 k=0 \\ \Rightarrow & h^2+2 h-2 k+2=0 \\ \Rightarrow & x^2+2 x-2 y+2=0 \\ \Rightarrow & 2 y=x^2+2 x+2 \\ \Rightarrow & y=\frac{1}{2}\left(x^2+2 x\right)+1 \\ \Rightarrow & y=\frac{1}{2}(x+1)^2-\frac{1}{2}+1 \\ \Rightarrow & y=\frac{1}{2}(x+1)^2+\frac{1}{2} \\ \Rightarrow & (x+1)^2=2\left(y-\frac{1}{2}\right) \end{array} $

Which represents parabola with focus $(-1,1)$

Given, equation of circles

$ \begin{aligned} & x^2+y^2+2 x-2 y+1=0 \text { and } \\ & x^2+y^2-2 x+2 y-2=0 \end{aligned} $

$\therefore$ Equation of circle passing through the point of intersection of two given circle is

$ \begin{array}{r} \left(x^2+y^2+2 x-2 y+1\right)+\lambda \left(x^2+y^2-2 x+2 y-2\right)=0 \\ \Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) x +(-2+2 \lambda) y+(1-2 \lambda)=0 \\ \Rightarrow \quad x^2+y^2+\frac{(2-2 \lambda)}{1+\lambda} x+\frac{(-2+2 \lambda)}{1+\lambda} y +\frac{1-2 \lambda}{1+\lambda}=0 \end{array} $

Coordinate of centre of circle

$ =(-g,-f)=\left(-\frac{1-\lambda}{1+\lambda},-\frac{-1+\lambda}{1+\lambda}\right) $

$ \begin{aligned} & \therefore \text { Radius of } \\ & \begin{aligned} &\left(-\frac{1-\lambda}{1+\lambda}\right)^2+\left(-\frac{-1+\lambda}{1+\lambda}\right)^2-\left(\frac{1-2 \lambda}{1+\lambda}\right)=14 \\ & \Rightarrow \quad \frac{(1-\lambda)^2}{(1+\lambda)^2}+\frac{(\lambda-1)^2}{(1+\lambda)^2}-\frac{1-2 \lambda}{1+\lambda}=14 \\ & \Rightarrow \quad(1-\lambda)^2+(\lambda-1)^2-(1+\lambda)(1-2 \lambda) \\ &=14(1+\lambda)^2 \\ & \Rightarrow \quad 2(\lambda-1)^2-\left(1-2 \lambda+\lambda-2 \lambda^2\right) \\ &=14\left(1+\lambda^2+2 \lambda\right) \\ & \Rightarrow \quad 2\left(\lambda^2+1-2 \lambda-\left(1-\lambda-2 \lambda^2\right)\right. \\ &=14\left(1+\lambda^2+2 \lambda\right) \\ & \Rightarrow \quad 4 \lambda^2+1-3 \lambda=14+14 \lambda^2+28 \lambda \\ & \Rightarrow \quad 10 \lambda^2+31 \lambda+13=0 \end{aligned} \end{aligned} $

$\Rightarrow$ This is a quadrate equation is $\lambda$ the existence of real values for $\lambda$ conforms that such circle exists.

Coordinates of centre

$ \begin{array}{ll} & h=\frac{\lambda-1}{1+\lambda}, k=-\frac{\lambda-1}{1+\lambda} \\ \Rightarrow & h=-k \Rightarrow h+k=0 \\ \Rightarrow & x+y=0 \end{array} $

$ \begin{aligned} &\text { Given equation of circle }\\ &x^2+y^2-14 x+6 y+33=0 \ldots \text { (i) } \end{aligned} $

$ \begin{aligned} & \text { Centre of circle }=(-g,-f) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =(7,-3) \\ & \text { and radius of circle }=\sqrt{g^2+f^2-c} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sqrt{49+9-33} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sqrt{25}=5 \end{aligned} $

Since, circle cuts the $X$-axis, $y=0$

$ \begin{array}{ll} & x^2-14 x+33=0 \\ \Rightarrow & x^2-11 x-3 x+33=0 \\ \Rightarrow & x(x-11)-3(x-11)=0 \\ \Rightarrow & (x-3)(x-11)=0 \\ \Rightarrow & x=3,11 \\ \therefore & A=(3,0) \text { and } B=(11,0) \end{array} $

Since, $C$ is the mid-point of $A B$

$ \begin{aligned} \therefore \quad C & =\left(\frac{3+11}{2}, \frac{0+0}{2}\right) \\ & =(7,0) \end{aligned} $

$\therefore$ Equation of line $L$, which passes through $C$ and having slope $(-1)$ is

$ \begin{aligned} & y-0=(-1)(x-7) \\ & \Rightarrow \quad y=-x+7 \\ & \Rightarrow \quad x+y-7=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Since, $L$ is the radical axis of the circles $s$ and $S^{\prime}\left(S-S^{\prime}=0\right)$

$ \begin{aligned} & \therefore \quad\left(x^2+y^2-14 x+6 y+33\right) -\left(x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}\right)=0 \\ & \Rightarrow\left(-14-2 g^{\prime}\right) x+\left(6-2 f^{\prime}\right) y+\left(33-c^{\prime}\right)=0 \end{aligned} $

Which is identical with line (ii)

$ \begin{array}{ll} \therefore & \frac{-14-2 g^{\prime}}{1}=\frac{6-2 f^{\prime}}{1}=\frac{33-c^{\prime}}{-7}=k \\ \Rightarrow & -14-2 g^{\prime}=k \text { and } 6-2 f^{\prime}=k \\ \therefore & -14-2 g^{\prime}=6-2 f^{\prime} \\ \Rightarrow & g^{\prime}-f^{\prime}=-10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Since, centre of $S^{\prime}$ lies on (ii)

$ \begin{aligned} \therefore & & -g^{\prime}-f^{\prime} & =7 \\ \Rightarrow & & g^{\prime}+f^{\prime} & =-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Solving Eqs. (iii) and (iv), we get

$ \begin{array}{ll} & g^{\prime}=\frac{-17}{2}, f^{\prime}=\frac{3}{2} \\ \therefore & -14-2\left(-\frac{17}{2}\right)=k \\ \Rightarrow & k=3 \end{array} $

Also, $\frac{33-c^{\prime}}{-7}=k=3$

$ \begin{array}{ll} \Rightarrow & 33-c^{\prime}=-21 \\ \Rightarrow & c^{\prime}=54 \end{array} $

$\therefore$ Equation of required circle is

$ \begin{aligned} & x^2+y^2+2\left(\frac{-17}{2}\right) x+2\left(\frac{3}{2}\right) y+54=0 \\ & \Rightarrow x^2+y^2-17 x+3 y+54=0 \end{aligned} $

Given equation is

$ a x^2+a y^2+2 g x+2 f y-2=0 $

Put $(-1,0)$ into this equation, we get

$ \begin{aligned} & a(-1)^2+a(0)^2+2 g(-1)+2 f(0)-2=0 \\ & \Rightarrow \quad a-2 g-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Put $(-1,1)$, we get

$ \begin{aligned} & a(-1)^2+a(1)^2+2 g(-1)+2 f(1)-2=0 \\ & \Rightarrow \quad a+a-2 g+2 f-2=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2 a-2 g+2 f-2=0 \\ & \Rightarrow \quad a-g+f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Put (1, 1), we get

$ \begin{aligned} & a(1)^2+a(1)^2+2 g(1)+2 f(1)-2=0 \\ & \Rightarrow \quad a+g+f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Subtract Eq. (ii) from Eq. (iii),

$ \begin{aligned} & (a+g+f-1)-(a-g+f-1)=0 \\ & \Rightarrow \quad 2 g=0 \\ & \Rightarrow \quad g=0 \end{aligned} $

Put $g=0$ into Eq. (i), we get

$ \begin{aligned} & & a-2(0)-2 & =0 \\ \Rightarrow & & a & =2 \end{aligned} $