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} $