Circle

Let $(h, k)$ be the mid-point of a chord of the circle $x^2+y^2=25$.

Then, equation of the chord is $h x+k y-25$

$ =h^2+k^2-25 \Rightarrow h x+k y=h^2+k^2 $

If this is parallel to $x-2 y+3=0$, then $-\frac{h}{k}=\frac{1}{2} \Rightarrow 2 h+k=0$

So, the locus of $(h, k)$ is $2 x+y=0$.

The point $(2 a, a+1)$ will be an interior point of the larger segment of the circle.

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

if the point $(2 a, a+1)$ is an interior point and the point $(2 a, a+1)$ and the centre $(1,1)$ are on the same side of the chord $x-y+1=0$.

$ \begin{aligned} & \therefore(2 a)^2+(a+1)^2-2(2 a)-2(a+1)-8 < 0 \\ & \text { and }(2 a-a-1+1)(1-1+1) > 0 \\ & \Rightarrow \quad 5 a^2-4 a-9 < 0 \text { and } a > 0 \\ & \Rightarrow \quad(5 a-9)(a+1)<0 \text { and } a > 0 \\ & \Rightarrow \quad-1 < a <\frac{9}{5} \text { and } a > 0 \\ & \Rightarrow \quad a \in\left(0, \frac{9}{5}\right) \end{aligned} $

Given $ x=a\left(\frac{1-t^{2}}{1+t^{2}}\right) $ and $ y=\frac{2 a t}{1+t^{2}} $

Let $ \quad t=\tan \theta $

$ \Rightarrow \quad x=a\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) $

and $ y=\frac{2 a \tan \theta}{1+\tan ^{2} \theta} $

$ \Rightarrow \quad x=a \cos 2 \theta $ and $ y=a \sin 2 \theta $

$ \Rightarrow \cos 2 \theta=\frac{x}{a} $ and $ \sin 2 \theta=\frac{y}{a} $

Squaring both sides, we get

$ \cos ^{2} 2 \theta=\frac{x^{2}}{a^{2}} \quad \ldots \text { (i) } $

and $ \sin ^{2} 2 \theta=\frac{y^{2}}{a^{2}} \quad \ldots \text { (ii) } $

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

$ \cos ^{2} 2 \theta+\sin ^{2} 2 \theta=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}} $

$ \Rightarrow \quad x^{2}+y^{2}=a^{2} $

$ \therefore $ Locus is a circle having centre at origin and radius ' $ a $ '.

According to question,

$ O C=\sqrt{h^{2}+k^{2}} $

In $ \triangle O C B $,

$ \cos 45^{\circ}=\frac{\sqrt{h^{2}+k^{2}}}{2} $

$ \Rightarrow \quad \frac{\sqrt{h^{2}+k^{2}}}{2}=\frac{1}{\sqrt{2}} $

$ \Rightarrow \quad h^{2}+k^{2}=2 $

Replacing $ h \rightarrow x $ and $ k \rightarrow y $

Locus $ \Rightarrow x^{2}+y^{2}=2 $

Solving the given equations,

$\begin{aligned} x^2+(m x+c)^2 & =a^2 \\ \Rightarrow \quad x^2+m^2 x^2+c^2+2 m x c & =a^2 \\ \Rightarrow \quad\left(1+m^2\right) x^2+2 m c x+c^2-a^2 & =0 \end{aligned}$

Since, the straight line touches the circle at a point.

So, the discriminant $=0$

$\begin{aligned} & \Rightarrow(2 m c)^2-4\left(1+m^2\right)\left(c^2-a^2\right)=0 \\ & \Rightarrow 4 m^2 c^2-4 c^2-4 m^2 c^2+4 a^2+4 a^2 m^2=0 \\ & \Rightarrow-4 c^2=-4 a^2\left(1+m^2\right) \Rightarrow c^2=a^2\left(1+m^2\right) \end{aligned}$

The normal line will makes $45^{\circ}$ angle with positive $\mathrm{X}$-axis because it is inclined at $45^{\circ}$ with the straight line $y=6$ which is parallel to the $\mathrm{X}$-axis.

$\therefore$ Slope of normal line $=m=\tan 45^{\circ}=1$

The centre of the circle, $x^2+y^2=25$ is $(0,0)$.

So, the normal line will passes through $(0,0)$, because normal line at any point $P$ on the circle passes through its centre.

Thus, the equation of normal line be $y=x$ and the point $P$ lies on the straight line $y=x$.

The equation of tangent to the given circle at $(\cos \theta, \sin \theta)$ be, $x \cos \theta+y \sin \theta=1$

It meets the $X$-axis at $(\sec \theta, 0)$ and $Y$-axis at $(0, \operatorname{cosec} \theta)$ Let $R(h, k)$ be the mid-point of $P Q$.

$\begin{gathered} \therefore \quad \frac{\sec \theta+0}{2}=h \text { and } \frac{0+\operatorname{cosec} \theta}{2}=k \\ \Rightarrow \quad \frac{1}{2 h}=\cos \theta \quad \text{... (i)} \end{gathered}$

and $\frac{1}{2 k}=\sin \theta \quad \text{.... (ii)}$

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

$ \frac{1}{4 h^2}+\frac{1}{4 k^2}=\cos ^2 \theta+\sin ^2 \theta $

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

$ \text { Hence, the required locus be } x^2+y^2-4 x^2 y^2=0 $

Let $P\left( {t,{{4t - 20} \over 5}} \right)$ be a point on the line $4x - 5y = 20$. Then the chord of contact of tangents drawn from P to the circle ${x^2} + {y^2} = 9$ is

$tx + \left( {{{4t - 20} \over 5}} \right)y = 9$ ...... (i)

Let $\theta$ (h, k) be the mid-point of this chord of contact, then its equation is also $hx + ky = {h^2} + {k^2}$ ..... (ii) [using $T = S'$]

Clearly Eqs. (i) and (ii) represent the same line

$\therefore$ ${t \over h} = {{4t - 20} \over {5k}} = {9 \over {{h^2} + {k^2}}}$

$ \Rightarrow {t \over h} = {9 \over {{h^2} + {k^2}}}$ and ${t \over h} = {{4t - 20} \over {5k}}$

$ \Rightarrow t = {{9h} \over {{h^2} + {k^2}}}$ and $t = {{20h} \over {4h - 5k}}$

$ \Rightarrow {{9h} \over {{h^2} + {k^2}}} = {{20h} \over {4h - 5k}}$

$ \Rightarrow h\{ 20({h^2} + {k^2}) - 36h + 45k\} = 0$

$x = 0$ or $[20({x^2} + {y^2}) - 36x + 45y] = 0$

Since, the circle touches X-axis

(x $-$ h)² + (y $-$ k)² = k² ..... (i)

Also, it passes through the points (1, $-$2) and (3, $-$4).

(1 $-$ h)² + ($-$2 $-$ k)² = k² ...... (ii)

and (3 $-$ h)² + ($-$4 $-$ k)² = k² ...... (iii)

Subtracting Eq. (iii) from Eq. (ii), we get

h = k + 5

On solving these equations, we get

By putting the values of (h, k) = ($-$5, $-$10) or (3, $-$2) in Eq. (i), we get

x² + y² + 10x + 20y + 25 = 0 or x² + y² $-$ 6x + 4y + 9 = 0