Q.The solution of the differential equation 𝑥𝑑𝑥 + 𝑦𝑑𝑦 = 0 represents a family of
(A) straight lines
(B) parabolas
(C) Circles
(D) Ellipses

Method 1 — Algebraic (substitute $z=x+iy$).

Setting them equal and squaring (both sides are non-negative, so squaring is safe):

The $y^{2}$ terms cancel:

So the locus is the line $x=0$ with $y$ free — that is the $y$-axis (the imaginary axis).

Method 2 — Geometric (the faster way to see it).

In the Argand plane, $|z-z_0|$ is the distance from $z$ to the point $z_0$. Here:

• $|z+1|=|z-(-1)|$ = distance of $z$ from $A(-1,0)$,
• $|z-1|$ = distance of $z$ from $B(1,0)$.

The condition "distance to $A$ = distance to $B$" defines the perpendicular bisector of $AB$. Since $A$ and $B$ are symmetric about the origin on the real axis, that bisector is the vertical line through the origin — the $y$-axis. ✓ Same conclusion. …

We are given the circle equation

and told it passes through the origin (which it does, since plugging (0,0) gives 0). The question asks for the equation of one of its diameters. A diameter is any line that goes through the center of the circle. So the problem reduces to: find the center, then check which given line passes through it.

1. Find the center by completing the square. Group the $x$ terms and $y$ terms:

Complete the square for $x$:

Complete the square for $y$:

Substitute back:

So the center is $(3, -1)$ and radius $\sqrt{10}$.

2. Check which line passes through the center.

   A diameter must contain the center. Test each option:

   • (A) $3x - y = 0$: at $(3, -1)$ → $3(3) - (-1) = 9 + 1 = 10 \neq 0$. No. …

We have a circle that:

• touches the $x$-axis (so the $x$-axis is tangent to the circle),
• passes through $(1,1)$,
• has its centre on the line $x+y=3$ in the first quadrant.

The key idea: “touches the $x$-axis” means the distance from the centre to the $x$-axis equals the radius. If the centre is $(h,k)$, then the distance to the $x$-axis is simply $|k|$. Since the centre is in the first quadrant, $k>0$, so the radius $r = k$.

Now we use the other conditions to find $h$ and $k$.

1. Centre lies on $x+y=3$

   So $h + k = 3$.

   Hence $h = 3 - k$.

2. Circle passes through $(1,1)$

   The distance from centre $(h,k)$ to $(1,1)$ equals the radius $k$.

Square both sides:

3. Substitute $h = 3 - k$

Expand:

4. Interpret the solutions
   • If $k=1$, then $h=2$. Centre $(2,1)$, radius $1$. This is in the first quadrant. …

Concept & Intuition

We have an ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. Its foci lie on the major axis (the x‑axis here, since 16 > 9). The circle’s center is at $(0,3)$, directly above the ellipse’s center. The circle passes through both foci, so the distance from $(0,3)$ to either focus is the radius. Once we find the foci coordinates, the radius is just the Euclidean distance; doubling gives the diameter.

1. Identify the ellipse parameters For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$, the foci are at $(\pm c,0)$ where $c^2 = a^2 - b^2$. Here $a^2 = 16$, $b^2 = 9$, so

Thus the foci are $F_1 = (-\sqrt{7},0)$ and $F_2 = (\sqrt{7},0)$.

2. Find the radius of the circle The circle’s center is $C = (0,3)$. The radius $r$ is the distance from $C$ to either focus (say $F_2$): $r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4.$ …

For the X-intercept, put $y=0$ in $x^2+y^2+3x-y+2=0$:

The intercept length is the distance between these points:

$|-1 - (-2)| = 1.$ …

Let the centre be $(h,k)$, radius $r$. Passing through $(0,0)$ and $(1,0)$:

Subtracting: $2h-1=0\Rightarrow h=\tfrac12$.

Since the circle passes through the origin, the distance from the centre to the origin equals $r$: $\sqrt{h^2+k^2}=r$. The fixed circle $x^2+y^2=9$ has centre $O$ and radius $3$, and the origin is inside it, so the small circle is internally tangent: …

[!TLDR]

The centre is the intersection of the two diameters, $(8,-2)$; with radius 10 this gives a circle of the form $x^2+y^2-16x+4y-32=0$, matching the radius-10 option (A).

Concept

Every diameter of a circle passes through its centre, so the centre is the point common to both diameter lines. Once the centre $(h,k)$ and radius $r$ are known, the circle is $(x-h)^2+(y-k)^2=r^2$ (CBSE/NCERT Class 11 conic sections / straight lines).

Solution

Solve the two diameters simultaneously:

Subtracting the first from the second: $(x+2y)-(x+y)=4-6$, so $y=-2$. Then $x=6-y=6-(-2)=8$. Centre $=(8,-2)$.

With $r=10$: …

Circle 1: $(x+1)^2+y^2=4$. Circle 2: $(x-1)^2+y^2=9$.

Subtract to eliminate $y^2$:

Substitute into Circle 1:

$(-1.25+1)^2 + y^2 = 4 \;\Rightarrow\; (-0.25)^2 + y^2 = 4 \;\Rightarrow\; y^2 = 4 - 0.0625 = 3.9375 = \frac{63}{16}.$ …

Concept: A point P lies inside a circle S = 0 iff S(P) < 0, and outside iff S(P) > 0.

S = x^2 + y^2 + 2x + 3y + 1 , at P(-3, -2):

S(P) = 9 + 4 + 2(-3) + 3(-2) + 1 = 9 + 4 - 6 - 6 + 1 = 2 > 0 -> P lies OUTSIDE S.

S' = x^2 + y^2 + 4x + 3y + 2 , at P(-3, -2): …

Concept: For the circle x^2 + y^2 + 2gx + 2fy + c = 0, the length of the intercept cut on the Y-axis is 2*sqrt(f^2 - c).

Here: x^2 + y^2 + 4x - 7y + 12 = 0

2g = 4 -> g = 2

2f = -7 -> f = -7/2

c = 12 …

For a circle $S\equiv x^2+y^2+2gx+2fy+c=0$, a point is inside if substituting it makes $S<0$ and outside if $S>0$.

For $S\equiv x^2+y^2-2x-4y-4$ at $(-2,-1)$:

For $S'\equiv x^2+y^2-4x-2y-16$ at $(-2,-1)$: …

Concept: Two circles intersect in two distinct points iff

|r1 - r2| < d < r1 + r2 , where d is the distance between centres.

Circle 1: (x-1)^2 + (y-3)^2 = r^2 -> C1 = (1, 3), radius r.

Circle 2: x^2 + y^2 - 8x + 2y + 8 = 0 -> C2 = (4, -1), radius = sqrt(16 + 1 - 8) = sqrt(9) = 3.

Distance between centres:

d = sqrt((4-1)^2 + (-1-3)^2) = sqrt(9 + 16) = 5 …