Q.Find the shortest distance of the point $(0, c)$ from the parabola $y = x^2$, where $\dfrac{1}{2} \le c \le 5$.

We are told that $(3, -2)$ is the centre of the circle

The general form of a circle is

with centre $(-g, -f)$. So we have

Thus the circle’s equation becomes

Completing the square:

so

Hence the centre is $C(3, -2)$ and radius $r = 6$.

Now, point $P(-1, -5)$ is given. The distance from $P$ to the centre $C$ is

Since $CP = 5 < r = 6$, point $P$ lies inside the circle.

Key idea: For a point inside a circle, the point on the circle closest to it lies on the line joining the centre to that point, in the direction from the centre toward the point. The distance from the centre to the closest point is just the radius, so the required point $A$ is on the ray from $C$ through $P$, at a distance $r$ from $C$.

1. Find the direction vector from centre $C(3, -2)$ to $P(-1, -5)$:

Its length is $5$, so the unit vector in that direction is

2. Move from centre $C$ along this direction by the radius $r = 6$: $A = C + 6 \cdot \hat{u} = (3, -2) + 6\left( -\frac{4}{5},\; -\frac{3}{5} \right) = \left( 3 - \frac{24}{5},\; -2 - \frac{18}{5} \right).$ …

We start with the concept: A circle that touches both coordinate axes must have its center at $(r, r)$ and radius $r$, because the distance from the center to each axis equals the radius. So the general equation is $(x - r)^2 + (y - r)^2 = r^2$. Two such circles will intersect at two points A and B. Given one intersection point A = (1,2), we can find the two possible radii, then the second intersection point, and finally the distance AB.

1. Set up the circle equation For a circle with center $(r, r)$ and radius $r$, the equation is:

Expanding:

Simplifies to:

2. Plug in point A = (1,2) Since A lies on the circle:

Solving: $(r - 1)(r - 5) = 0$, so $r = 1$ or $r = 5$.

3. Identify the two circles

   • Circle 1: center $(1,1)$, radius 1. Equation: $(x-1)^2 + (y-1)^2 = 1$.
   • Circle 2: center $(5,5)$, radius 5. Equation: $(x-5)^2 + (y-5)^2 = 25$.

4. Find the line through the intersection points

   Subtract the equations of the two circles to get the radical axis (the line containing both intersection points).

   Circle 1 expanded: $x^2 + y^2 - 2x - 2y + 1 = 0$.

   Circle 2 expanded: $x^2 + y^2 - 10x - 10y + 25 = 0$.

   Subtract: $(x^2+y^2 - 2x - 2y + 1) - (x^2+y^2 - 10x - 10y + 25) = 0$

So the line AB is $x + y = 3$.

5. Find the second intersection point B We know A = (1,2) lies on this line (since $1+2=3$). To find B, solve the system of the line and one circle (say circle 1). From $x + y = 3$, we have $y = 3 - x$. Substitute into circle 1: $(x-1)^2 + ((3-x)-1)^2 = 1$ …

The key idea: A circle that touches both coordinate axes must have its center on the line $y=x$ (since the distances to both axes are equal). If the radius is $r$, the center is $(r,r)$. Two such circles intersect at $A(1,2)$ and $B$. The line joining their centers is perpendicular to the common chord $AB$, and the chord length can be found using the geometry of intersecting circles.

1. Equation of a circle touching both axes For a circle with center $(r,r)$ and radius $r$, the equation is

Expanding:

2. Both circles pass through $A(1,2)$ Substitute $(1,2)$ into the equation:

So $r^2 - 6r + 5 = 0$, giving $r = 1$ or $r = 5$.

Thus the two circles have radii $1$ and $5$, with centers $C_1(1,1)$ and $C_2(5,5)$.

3. Geometry of the common chord The line joining the centers $C_1C_2$ is along $y=x$. The common chord $AB$ is perpendicular to this line. The distance between centers is

4. Length of the common chord For two intersecting circles of radii $r_1=1$, $r_2=5$ with center distance $d$, the distance from $C_1$ to the chord (the perpendicular from center to chord) is given by $$$$

Concept & Intuition

We have a circle with known centre $C(3,-2)$ and an equation that still contains unknown $g$ and $f$. But the centre tells us $g$ and $f$ directly: for a circle $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$. So we can find $g$ and $f$, then the radius.

The point on the circle that is closest to an external point $P$ lies on the line joining the centre $C$ to $P$. Why? Because the shortest distance from a point to a circle is along the radial line: the distance from $P$ to any point on the circle is minimized when that point is the intersection of the line $CP$ with the circle, on the side nearer to $P$. So we find that intersection.

Step-by-step solution

1. Find $g$ and $f$ from the centre The circle is $S \equiv x^2 + y^2 + 2gx + 2fy - 23 = 0$. Centre is $(-g, -f)$. Given centre $(3, -2)$, we have

2. Find the radius Radius $r = \sqrt{g^2 + f^2 - c}$ where $c = -23$.

3. Equation of line through centre $C(3,-2)$ and point $P(-1,-5)$ Slope $m = \frac{-5 - (-2)}{-1 - 3} = \frac{-3}{-4} = \frac{3}{4}$. Using point-slope form with $C$:

Multiply by 4: $4y + 8 = 3x - 9$ → $3x - 4y = 17$.

4. Find intersection of this line with the circle Circle equation: $x^2 + y^2 - 6x + 4y - 23 = 0$ (substituting $g=-3, f=2$). From line: $x = \frac{17 + 4y}{3}$. Substitute into circle:

Multiply by 9:

Expand:

Combine: $25y^2 + (136 - 72 + 36)y + (289 - 306 - 207) = 0$

Divide by 1 (or simplify by 1): actually divide by 1? Let's divide by 1:

Divide by 1? Better divide by 1? Actually divide by 1 is trivial. Let's solve:

So $y = \frac{80}{50} = \frac{8}{5}$ or $y = \frac{-280}{50} = -\frac{28}{5}$.

5. Choose the point closer to $P$ The two intersection points are on opposite sides of $C$. The one nearer to $P$ will have the smaller distance to $P$. For $y = \frac{8}{5}$, $x = \frac{17 + 4(8/5)}{3} = \frac{17 + 32/5}{3} = \frac{(85+32)/5}{3} = \frac{117}{15} = \frac{39}{5}$. For $y = -\frac{28}{5}$, $x = \frac{17 + 4(-28/5)}{3} = \frac{17 - 112/5}{3} = \frac{(85-112)/5}{3} = \frac{-27}{15} = -\frac{9}{5}$. Point $P$ is $(-1,-5)$. Compare distances:
   • $(\frac{39}{5}, \frac{8}{5})$: difference in x ≈ 8.8, y ≈ 6.6 → large.
   • $(-\frac{9}{5}, -\frac{28}{5})$: difference in x = $-1.8 + 1 = -0.8$, y = $-5.6 + 5 = -0.6$ → much smaller. …

We start with $z = (\alpha + i\beta)(2 + 7i)$. Multiplying out:

For $z$ to be purely imaginary, the real part must be zero:

Since $\alpha, \beta$ are non-zero integers, the smallest positive integer solution is $\alpha = 7, \beta = 2$ (or their negatives). This is the fundamental Diophantine condition.

Now, $|z|^2$ is the square of the modulus:

But the first term is zero by our condition, so:

Using $2\alpha = 7\beta$, we can express $\alpha = \frac{7\beta}{2}$. Since $\alpha$ must be an integer, $\beta$ must be even. Let $\beta = 2k$, then $\alpha = 7k$, where $k$ is a non-zero integer. Substitute into $|z|^2$:

Thus:

$|z|^2 = (53k)^2 = 2809 k^2.$ …