Q.Show that the complex number $z$, satisfying the condition $\arg\left(\dfrac{z-1}{z+1}\right)=\dfrac{\pi}{4}$ lies on a circle.

Rewriting $x^{2}+y^{2}+6x-8y-24=0$ as $(x+3)^{2}+(y-4)^{2}=9+16+24=49$, the radius is $r=7$.

A chord subtending angle $2\alpha=60^\circ$ at the centre has length $2r\sin\alpha=2r\sin30^\circ$:

$L=2(7)\sin30^\circ=14\cdot\tfrac12=7\text{ units}.$ …

With $y=0$: $x^2-6x-55=0$.

Discriminant $=36+220=256$, $\sqrt{256}=16$, so $x=\frac{6\pm16}{2}=11$ or $-5$. …

Every diameter passes through the centre, hence the centre lies on both lines. Multiply the first by $3$ and the second by $2$:

$6x-9y=15,\qquad 6x-8y=14.$ …

The circles have centers $(2,3)$ (radius $3$) and $(2,-3)$ (radius $a$). The distance between centers is

For two distinct intersection points: …

The point $(2,k)$ lies on the circle $(x-2)^2+(y+1)^2=4$, so

$(2-2)^2+(k+1)^2=4,$ …

From the parametric equations,

Using $\cos^2\theta+\sin^2\theta=1$: …

Concept and Intuition

$PQ$ is a diameter of the second circle, so its midpoint is that circle's centre and $|PQ|=2\times2=4$. As a chord of the first circle, use the perpendicular-from-centre relation.

Step-by-Step Solution

1. Second circle: centre $(1,3)$, radius $2$; so midpoint of $PQ=(1,3)$ and $|PQ|=4$, half-chord $=2$.
2. First circle centre $(2,1)$. Distance to midpoint $d=\sqrt{(2-1)^2+(1-3)^2}=\sqrt5$. …

Concept and Intuition

$C:(x-3)^2+(y+1)^2=10$ has centre $(3,-1)$ and radius $\sqrt{10}$. Check each claim against these facts.

Step-by-Step Solution

1. Radius $=\sqrt{9+1}=\sqrt{10}$ — (A) correct.
2. $(3,-1)$ is the centre, hence inside — (B) correct.
3. $S(7,3)=49+9-42+6=22>0$ — outside — (C) correct. …

[!TLDR]

The points lie on the rectangular hyperbola $xy=1$; intersecting with a circle yields a quartic in $x$ whose product of roots is $1$, so the product of the four cosines is $1$.

Concept

A circle meets a rectangular hyperbola in up to four points. If we substitute the hyperbola's relation into the circle's equation, the resulting quartic's roots are the $x$-coordinates of the intersection points, and their product follows from Vieta's formulas — coordinate-geometry ideas in the NCERT/CBSE-aligned KEAM syllabus.

Solution

Let $x=\cos\theta_i$ and $y=\sec\theta_i=\dfrac{1}{\cos\theta_i}=\dfrac{1}{x}$. Every point $(\cos\theta_i,\sec\theta_i)$ therefore satisfies

i.e. all four lie on the rectangular hyperbola $xy=1$.

A general circle has equation

Substitute $y=\dfrac{1}{x}$:

$x^2+\frac{1}{x^2}+2gx+\frac{2f}{x}+k=0.$ …

Concept and Intuition

Substitute the line into the circle to get a quadratic in $x$; its roots give the $x$-coordinates of intersection.

Step-by-Step Solution

1. $y=x+2$ into $(x-2)^2+y^2=16$: $(x-2)^2+(x+2)^2=16$.
2. Expand: $x^2-4x+4+x^2+4x+4=16\Rightarrow 2x^2+8=16$.
3. $x^2=4\Rightarrow x=\pm2$. …

Concept and Intuition

The tangent to $(x-h)^2+(y-k)^2=r^2$ at a point $(x_1,y_1)$ on it is $(x-h)(x_1-h)+(y-k)(y_1-k)=r^2$.

Step-by-Step Solution

1. Here $h=5,k=0,r^2=25$, point $(2,4)$ (check: $9+16=25$ ✓).
2. Tangent: $(x-5)(2-5)+(y)(4)=25 \Rightarrow -3(x-5)+4y=25$.
3. Expand: $-3x+15+4y=25 \Rightarrow -3x+4y-10=0 \Rightarrow 3x-4y+10=0$. …