Circle

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 $\text{Radius} = \frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}.$

Consider centre as $\mathrm{P}(0, \alpha), \alpha>0$

$\left|\frac{2(0)-\alpha}{\sqrt{5}}\right|=\mathrm{r}$

$\begin{aligned} & |-\alpha|=\sqrt{5} r \\ & \alpha=\sqrt{5} r \\ & \therefore \alpha+r=5+\sqrt{5} \\ & \sqrt{5} r+r=\sqrt{5}(\sqrt{5}+1) \\ & r=\sqrt{5}, \alpha=5 \\ & \therefore P(0,5) \end{aligned}$

Foot of perpendicular from $P$ to line $2 x-y=0$

$\begin{aligned} & \frac{x-0}{2}=\frac{y-5}{-1}=\frac{-(2(0)-5)}{5}=1 \\ & x=2, y=4 \quad A_1(2,4) \end{aligned}$

Let $\mathrm{B}(\mathrm{p}, \mathrm{q}) \quad \therefore \frac{\mathrm{p}+2}{2}=0, \frac{\mathrm{q}+4}{2}=5$

$\therefore \mathrm{p}=-2, \mathrm{q}=6 \quad B(-2,6)$

Here if we add center of circles G₁, G₂, G₃ ....... G_n, then we get a polygon of n sides.

From figure you can see one side of polygon makes angle $\theta$ with the center.

$\therefore$ n sides make angle = n$\theta$

We know, $n\theta = 2\pi $

$ \Rightarrow \theta = {{2\pi } \over n}$

Here triangle OMN is an isosceles triangle. Line joining of point O and midpoint O of MN (point A) is perpendicular to line MN and perpendicular bisector of angle $\theta$.

$\therefore$ $\angle MOA = {\theta \over 2} = {\pi \over n}$

From right angle triangle OMA,

$\sin {\pi \over n} = {r \over {R + r}}$

$ \Rightarrow R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over n}$

Option A:

When $n = 4$,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 4}$

$ \Rightarrow R + r = \sqrt 2 r$

$ \Rightarrow R = \left( {\sqrt 2 - 1} \right)r$

$\therefore$ Option (A) is wrong.

Option B :

When $n = 5$,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 5} = r{\mathop{\rm cosec}\nolimits} 36^\circ $

We know, in 0 to ${\pi \over 2}$, ${\mathop{\rm cosec}\nolimits} \theta $ is decreasing.

$\therefore$ ${\mathop{\rm cosec}\nolimits} 45^\circ < {\mathop{\rm cosec}\nolimits} 36^\circ < {\mathop{\rm cosec}\nolimits} 30^\circ $

$ \Rightarrow \sqrt 2 < {\mathop{\rm cosec}\nolimits} 36^\circ < 2$

$ \Rightarrow \sqrt 2 r < r{\mathop{\rm cosec}\nolimits} 36^\circ < 2r$

$ \Rightarrow \sqrt 2 r - r < r{\mathop{\rm cosec}\nolimits} 36^\circ - r < 2r - r$

$ \Rightarrow \left( {\sqrt 2 - 1} \right)r < R < r$

$\therefore$ $R < r$, when $n = 5$

$\therefore$ Option (B) is wrong.

Option C :

When $n = 8$,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 8}$

${\mathop{\rm cosec}\nolimits} \theta$ is decreasing function in 0 to ${\pi \over 2}$.

$\therefore$ ${\mathop{\rm cosec}\nolimits} {\pi \over 8} > {\mathop{\rm cosec}\nolimits} {\pi \over 4}$

$ \Rightarrow r{\mathop{\rm cosec}\nolimits} {\pi \over 8} > \sqrt 2 r$

$ \Rightarrow R + r > \sqrt 2 r$

$ \Rightarrow R > \left( {\sqrt 2 - 1} \right)r$

$\therefore$ Option (C) is correct.

Option D :

When $n = 12$, then

$R + r = r\left( {\cos ec{\pi \over {12}}} \right)$

$ \Rightarrow R + r = r \times {{2\sqrt 2 } \over {\sqrt 3 - 1}}$ [as $\cos ec{\pi \over {12}} = {{2\sqrt 2 } \over {\sqrt 3 - 1}}$]

$ \Rightarrow R + r = r \times {{2\sqrt 2 \left( {\sqrt 3 + 1} \right)} \over 2}$

$ \Rightarrow R + r = \sqrt 2 r\left( {\sqrt 3 + 1} \right)$

$ \Rightarrow R = \left[ {\sqrt 2 \left( {\sqrt 3 + 1} \right) - 1} \right]r$

$ \Rightarrow R < \sqrt 2 \left( {\sqrt 3 + 1} \right)r$

$\therefore$ Option (D) is correct.

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).

From the given data, the centre of the circle is C(3, 2).

Since, CA and CB are perpendicular to PA and PB, CP is the diameter of the circumcircle of triangle PAB. Its equation is

$(x - 3)(x - 1) + (y - 2)(y - 8) = 0$

or ${x^2} + {y^2} - 4x - 10y + 19 = 0$.

Equation of circle C is

${(x + 3)^2} + {(y - 5)^2} = 9 + 25 - 30 = 4$

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

Centre = (3, $-$5)

If L₁ is diameter, then it passes through the center (3, $-$5),

$ \therefore $ $2(3) + 3( - 5) + p - 3 = 0 \Rightarrow p = 12$

$\therefore$ L₁ is $2x + 3y + 9 = 0$

and L₂ is $2x + 3y + 15 = 0$

Distance of centre of circle from ${L_2}(d) = \left| {{{2(3) + 3( - 5) + 15} \over {\sqrt {{2^2} + {3^2}} }}} \right| = {6 \over {\sqrt {12} }} < 2$ [radius of circle]

$\therefore$ L₂ is a chord of circle C.

Statement 2 is false.

Slope of line $PQ = - \sqrt 3 $

Therefore, PQ make 120$^\circ$ angle with x-axis. So, side PR lies along X-axis.

Therefore, $F = \left( {\sqrt 3 ,0} \right)$

Now, equation CE is ${{x - \sqrt 3 } \over {{{ - \sqrt 3 } \over 2}}} = {{y - 1} \over {{1 \over 2}}} = 1$

$E = \left( {{{\sqrt 3 } \over 2},{3 \over 2}} \right)$

Equation of PR is X-axis, that is $y = 0$ and the equation of side QR is

$\left( {y - {3 \over 2}} \right) = \sqrt 3 \left( {x - {{\sqrt 3 } \over 2}} \right)$

$y = \sqrt 3 x$

Let centre of circle C be (h, k) then,

$\left| {{{\sqrt 3 h + k - 6} \over {\sqrt {3 + 1} }}} \right| = 1$

$\sqrt 3 h + k - 6 = 2, - 2$

$\sqrt 3 h + k = 4$ (rejecting 2 because origin and centre of point $\left( {\sqrt 3 ,1} \right)$, satisfies eq.)

eq. of circle (C) is

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

Clearly point E and F satisfy the equation in given option D.

As ${\left( {x - 2\sqrt 3 } \right)^2} + {(y - 1)^2} = 1$ not possible.

Area = $\frac{1}{2}$ (sum of parallel sides height)

$18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6$

Line, $y=\frac{-2 r}{\alpha}(x-2 \alpha)$ is tangent to circle $(x-r)^{2}+(y-r)^{2}=r^{2}$

$2 \alpha=3 r$ and $\alpha r=6$

$\Rightarrow r=2$

(A) $\to (p),(q)$

It is clear from the fig, that two intersecting circles have a common tangent and a common normal joining the centres.

(B) $\to (p),(q)$

(C) $\to (q),(r)$

Two circles when one is completely inside the other have a common normal C$_1$ C$_2$ but not common tangent.

(D) $\to (q),(r)$

Two branches of hyperbola have no common tangent but have a common normal joining S$_1$ and S$_2$.

Locus of the points of intersections of perpendicular tangents to the circles

${x^2} + {y^2} = {a^2}$

${x^2} + {y^2} = 2{a^2}$

$\therefore$ director circle of ${x^2} + {y^2} = 169$ is the circle of ${x^2} + {y^2} = (169)(2) = 338$

The point (17, 7) lies of on the circle ${x^2} + {y^2} = 338$. Thus, the tangent drawn from (17, 7) to the circle ${x^2} + {y^2} = 169$ are perpendicular.

$ \text { Here, } \mathrm{MP}=r_1+k $

$ \begin{aligned} & \Rightarrow \quad(h-0)^2+(k-b)^2=\left(r_1+k\right)^2 \\ & \Rightarrow \quad h^2+(k-b)^2=\left(r_1+k\right)^2 \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & h^2+k^2+b^2-2 b k & =r_1+k^2+2 r_1 k \\ \Rightarrow & h^2 & =-b^2+2 k\left(r_1+b\right) \end{array} $

Replace $h$ by $x$ and $k$ by $y$

$ x^2=-b^2+2 y\left(r_1+b\right) $

$\Rightarrow \quad$ Parabola.

$ \text { Diagonal of square with side length } 2 \text { is } 2 \sqrt{2} $

$ \begin{aligned} &\begin{aligned} \therefore & & \mathrm{OA} & =\sqrt{2} \\ & & \mathrm{AT}_1 & =\frac{\mathrm{OA}}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \end{aligned}\\ &\text { Area of } \Delta T_1 T_2 T_3 \end{aligned} $

                     $ \begin{aligned} & =\frac{1}{2}\left(\mathrm{AT}_1\right) \times\left(\mathrm{T}_2 \mathrm{~T}_3\right) \\ & =\frac{1}{2}\left(\mathrm{AT}_1\right) \times 2 \sqrt{2} \\ & =\frac{2 \sqrt{2}}{2 \sqrt{2}}=1 \end{aligned} $

let A, B and C be the centres of circles respectively.

We know,

AP, BP and CP bisects the angle formed by the sector at centre A

$\mathrm{P}$ is the point of incentre of $\triangle \mathrm{ABC}$ and therefore

$\begin{aligned} r & =\frac{\Delta}{s}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s} \\ & =\sqrt{\frac{(s-a)(s-b)(s-c)}{\mathrm{s}}} \end{aligned}$

Now, $2 s=7+8+9$ (from figure)

$\Rightarrow \quad s=12$

Now, $r=\sqrt{\frac{(12-7)(12-8)(12-9)}{12}}$

$=\sqrt{\frac{60}{12}}=\sqrt{5} \text { units }$