Circle

Let, P $\equiv$ (cos$\theta$, sin$\theta$)

$\therefore$ equation of tangent and normal at P

x cos$\theta$ + y sin$\theta$ = 1 ..... (1)

and y = x tan$\theta$ ...... (2)

Now, equation of tangent at S : x = 1 ...... (3)

Solving (1) and (3), Q $\equiv$ (1, cosec$\theta$ $-$ cot$\theta$)

$\therefore$ equation of straight line parallel to RS drawn from Q

y = cosec$\theta$ $-$ cot$\theta$ ..... (4)

Let, E $\equiv$ (h, k)

$\therefore$ k = h tan$\theta$ [from (2)]

or, $\tan \theta = {k \over h}$

Again, k = cosec$\theta$ $-$ cot$\theta$ [from (4)]

or, $k = {{1 - \cos \theta } \over {\sin \theta }}$

or, $k = {{1 - {h \over {\sqrt {{h^2} + {k^2}} }}} \over {{k \over {\sqrt {{h^2} + {k^2}} }}}} = {{\sqrt {{h^2} + {k^2}} - h} \over k}$

or, ${k^2} = \sqrt {{h^2} + {k^2}} - h$

or, $h + {k^2} = \sqrt {{h^2} + {k^2}} $

$\therefore$ locus of E $x + {y^2} = \sqrt {{x^2} + {y^2}} $

Clearly, points $\left( {{1 \over 3},{1 \over {\sqrt 3 }}} \right)$ and $\left( {{1 \over 3}, - {1 \over {\sqrt 3 }}} \right)$ are on locus of E.

Therefore, (A) and (C) are the correct options.

Let, the equation of the required circle is

${x^2} + {y^2} + 2gx + 2fy + c = 0$ ..... (1)

Circle (I) cuts the circle ${(x - 1)^2} + {y^2} = 16$

i.e., ${x^2} + {y^2} - 2x = 15$ orthogonally

$\therefore$ $2( - g + 0) = - 15 + c$

or, $ - 2g = - 15 + c$

The circle (1) also cuts the circle ${x^2} + {y^2} = 1$ orthogonally.

$\therefore$ 0 = $-$1 + c or, c = 1

$\therefore$ g = 7

Now, the circle (1) passes through the point (0, 1).

$\therefore$ $2f + 1 + c = 0$ or, $2f + 1 + 1 = 0$ or, f = $-$1

$\therefore$ the equation of the required circle is

${x^2} + {y^2} + 14x - 2y + 1 = 0$

whose centre is ($-$7, 1) and radius $ = \sqrt {49 + 1 - 1} = 7$ units

Therefore, (B) and (C) are the correct option.

Note :

The condition of the circle ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ cuts orthogonally to the circle ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ is $2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$

Let a circle $x^2+y^2+2 g x+2 f y+c=0$ touches $x$-axis and have $y$-intercept of length $2 \sqrt{7}$.

$\begin{aligned} & \therefore 2 \sqrt{g^2-c}=0 \text { and } 2 \sqrt{f^2-c}=2 \sqrt{7} \\ & \Rightarrow g^2=c \text { and } f^2-c=7 \\ & \Rightarrow c=g^2=f^2-7 \quad \text{... (i)} \end{aligned}$

Given, the circle touches $x$-axis at a distance 3 unit from origin.

$\begin{aligned} & \therefore|-g|=3 \\ & \Rightarrow g= \pm 3 \end{aligned}$

Put $g= \pm 3$ in the equation

$\Rightarrow c=9 \text { and } f= \pm 4$

Hence, the possible equation of circles are

$\begin{aligned} & x^2+y^2-6 x+8 y+9=0, \\ & x^2+y^2-6 x-8 y+9=0, \end{aligned}$

$\begin{aligned} & x^2+y^2+6 x+8 y+9=0 \text { and } x^2+y^2+6 x-8 y \\ & +9=0 \end{aligned}$

Hints:

The length of X-intercept and Y-intercept of the circle $x^2+y^2+2 g x+2 f y+c=0$ are $2 \sqrt{g^2-c}$ and $2 \sqrt{f^2-c}$ respectively.

Equation of tangent PT of the circle $x^2+y^2=4$ at $\mathrm{P}(\sqrt{3}, 1)$ is

$\Rightarrow \quad \sqrt{3} x+y=4$

Given, L is a line perpendicular to PT

$\therefore \quad \mathrm{L} \equiv x-\sqrt{3} y=\lambda$

Also given L is the tangent of circle

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

$\begin{array}{ll} \Rightarrow & 1=\frac{|3-\sqrt{3} \cdot 0-\lambda|}{\sqrt{1^2+(-\sqrt{3})^2}} \\ \Rightarrow & |3-\lambda|=2 \\ \Rightarrow & 3-\lambda= \pm 2 \\ \Rightarrow & \lambda=1,5 \end{array}$

Hence, the possible equation of line L are $x-\sqrt{3} y=1$ and $x-\sqrt{3} y=5$.

The equation of tangent of the circle $x^2+y^2=4$ is $y=m x \pm 2 \sqrt{1+m^2}\quad \text{.... (i)}$

Let $y=m x \pm 2 \sqrt{1+m^2}$ also touches $(x-3)^2+ y^2=1$

$\Rightarrow(x-3)^2+\left(m x \pm 2 \sqrt{1+m^2}\right)^2=1$

$\Rightarrow x^2-6 x+9+m^2 x^2+4\left(1+m^2\right) \pm 4 m \sqrt{1+m^2} x=1$

$\Rightarrow\left(1+m^2\right) x^2+\left(-6 \pm 4 m \sqrt{1+m^2}\right) x+4\left(m^2+3\right)=0$

Apply

$\begin{aligned} & \quad \left(-6 \pm 4 m \sqrt{1+m^2}\right)^2-4\left(1+m^2\right) \cdot 4\left(m^2+3\right)=0 \\ \Rightarrow \quad & 36+16 m^2\left(1+m^2\right) \pm 48 m \sqrt{1+m^2} \\ & \quad -16\left(m^4+4 m^2+3\right)=0 \\ \Rightarrow \quad & 4 m^2+1= \pm 4 m \sqrt{1+m^2} \end{aligned}$

On squaring both side

$\begin{array}{rlrl} \Rightarrow & 16 m^4+1+8 m^2 =16 m^2+16 m^4 \\ \Rightarrow & m^2 =\frac{1}{8} \\ \Rightarrow & m = \pm \frac{1}{2 \sqrt{2}} \end{array}$

$\begin{array}{ll} \text { Put } & m= \pm \frac{1}{2 \sqrt{2}} \text { in the equation (i) } \\ \Rightarrow & y= \pm \frac{x}{2 \sqrt{2}} \pm \frac{6}{2 \sqrt{2}} \\ \Rightarrow & 2 \sqrt{2} y= \pm x \pm 6 \end{array}$

Hence, the equation of common tangent of given circles are $2 \sqrt{2} y=-x+6,2 \sqrt{2} y=x+6, 2 \sqrt{2} y=-x-6$ and $2 \sqrt{2} y=x-6$.

Equation of the chord AB is $\mathrm{T}=0$

$\Rightarrow \alpha x+\left(\frac{4 \alpha-20}{5}\right) y=9\quad \text{... (i)}$

Also, equation of the chord AB whose mid-point is $(h, k)$ is $\mathrm{T}=\mathrm{S}_1$

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

$\because$ Equations (i) and (ii) both represent the same line

$\begin{aligned} & \therefore \frac{\alpha}{h}=\frac{\frac{4 \alpha-20}{5}}{k}=\frac{9}{h^2+k^2} \\ & \Rightarrow \alpha=\frac{9 h}{h^2+k^2} \quad \text { and } \quad \frac{4 \alpha-20}{5}=\frac{9 k}{h^2+k^2} \end{aligned}$

$\begin{aligned} & \Rightarrow \alpha=\frac{9 h}{h^2+k^2} \quad \text { and } \quad 4 \alpha=\frac{45 k}{h^2+k^2}+20 \\ & \Rightarrow \alpha=\frac{9 h}{h^2+k^2} \quad \text { and } \quad \alpha=\frac{45 k+20\left(h^2+k^2\right)}{4\left(h^2+k^2\right)} \end{aligned}$

$\begin{array}{ll} \Rightarrow & \frac{9 h}{h^2+k^2}=\frac{45 k+20\left(h^2+k^2\right)}{4\left(h^2+k^2\right)} \\ \Rightarrow & 36 h=45 k+20\left(h^2+k^2\right) \end{array}$

$\Rightarrow 20\left(x^2+y^2\right)-36 x+45 y=0$, which is the required locus.

Circle touching y-axis at (0, 2) is ${(x - 0)^2} + {(y - 2)^2} + \lambda x = 0$ passes through ($-$1, 0).

Therefore, $1 + 4 - \lambda = 0 \Rightarrow \lambda = 5$

Hence, ${x^2} + {y^2} + 5x - 4y + 4 = 0$

Substituting $y = 0 \Rightarrow x = - 1,\, - 4$.

Hence, the Circle passes through ($-$4, 0).