Q.Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line through the points $(2, 5)$ and $(-3, 6)$.

Normal to the plane. The line joins $(9,7,5)$ and $(2,10,0)$, so its direction is

and this is perpendicular to $\pi$.

Equation of $\pi$ through $(200,30,116)$ with normal $(-7,3,-5)$:

Axis intercepts. …

The point $Q$ lies on the Y-axis $(x=0)$ and on the line $5x-2y+6=0$:

$PQ$ is a tangent from the external point $Q$ to the circle

$S:\;x^2+y^2+6x+6y-2=0.$ …

Find $m:n$. The point dividing $A(-2,3),B(3,-2)$ in ratio $m:n$ is

It lies on $2x-3y+4=0$:

so $m:n=9:16$.

Required ratio. $-4m:3n=-4(9):3(16)=-36:48=-3:4.$ …

The given pair $3x^2 - 4xy + 5y^2 = 0$ represents two straight lines passing through the origin. Their slopes are the roots of the quadratic in $m$ obtained by dividing by $x^2$ (assuming $x \neq 0$): $5m^2 - 4m + 3 = 0$. Let these slopes be $m_1$ and $m_2$.

We need the pair of lines through $(2,3)$ that are perpendicular to these lines. If a line has slope $m$, a line perpendicular to it has slope $-\frac{1}{m}$. So the required slopes are $-\frac{1}{m_1}$ and $-\frac{1}{m_2}$.

The equation of a line through $(2,3)$ with slope $m'$ is $y - 3 = m'(x - 2)$. So the combined equation of the two required lines is:

where $m'_1 = -\frac{1}{m_1}$ and $m'_2 = -\frac{1}{m_2}$.

Now, the product $m'_1 m'_2 = \left(-\frac{1}{m_1}\right)\left(-\frac{1}{m_2}\right) = \frac{1}{m_1 m_2}$.

From the original quadratic $5m^2 - 4m + 3 = 0$, we have $m_1 m_2 = \frac{3}{5}$. So $m'_1 m'_2 = \frac{5}{3}$.

Also, the sum $m'_1 + m'_2 = -\left(\frac{1}{m_1} + \frac{1}{m_2}\right) = -\frac{m_1 + m_2}{m_1 m_2}$.

From the original quadratic, $m_1 + m_2 = \frac{4}{5}$. So $m'_1 + m'_2 = -\frac{4/5}{3/5} = -\frac{4}{3}$.

Now expand the product equation. Let $X = x-2$ and $Y = y-3$ for a moment. Then the equation is:

Substitute back $X = x-2$, $Y = y-3$:

Multiply through by 3 to clear denominators:

Now expand each term:

• $3(y^2 - 6y + 9) = 3y^2 - 18y + 27$
• $4(xy - 3x - 2y + 6) = 4xy - 12x - 8y + 24$ …

Let’s start with the concept. Two circles are orthogonal if their tangents at the point of intersection are perpendicular. This translates into a neat algebraic condition: for circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$, orthogonality means

Our unknown circle passes through the origin and has centre $(h, k)$. So its equation is

where $c$ is the constant term. Since it passes through $(0,0)$, plugging in gives $c = 0$. So the circle is simply

Here $g = -h$, $f = -k$, and $c = 0$.

Now we apply the orthogonality condition with each given circle.

1. First given circle: $x^2 + y^2 + 4x + 6y + 12 = 0$. Here $g_1 = 2$, $f_1 = 3$, $c_1 = 12$. Orthogonality with our circle ($g = -h$, $f = -k$, $c = 0$) gives:

Multiply by $-1$:

2. Second given circle: $x^2 + y^2 + 4x - 6y + 9 = 0$. Here $g_2 = 2$, $f_2 = -3$, $c_2 = 9$. Orthogonality gives: …

Concept and Intuition

Imagine a square. It has two diagonals. A key property of a square's diagonals is that they are equal in length, perpendicular to each other, and bisect each other at the center of the square. This center point is equidistant from all four vertices.

We are given two vertices, say $A$ and $B$, and the equation of one diagonal, $L$. The first crucial step is to figure out if $A$ and $B$ are adjacent vertices (forming a side) or opposite vertices (forming a diagonal).

1. Check if $A$ or $B$ lie on $L$: If either $A$ or $B$ lies on $L$, then $A$ and $B$ must be adjacent vertices, and $L$ is the diagonal that passes through one of them.
2. If neither $A$ nor $B$ lies on $L$: This means $A$ and $B$ must be the other two vertices, which are opposite to each other. In this case, the line segment $AB$ forms the other diagonal of the square. This is a much simpler scenario to work with!

Let's assume the second case for a moment. If $A$ and $B$ are opposite vertices, then the midpoint of $AB$ is the center of the square. This center must lie on both diagonals. So, if the midpoint of $AB$ lies on $L$, our assumption is correct.

Once we know the center of the square and the length of one diagonal (which is $AB$), we can determine the distance from the center to any vertex. The vertices we're looking for must lie on the line $L$ and be at this specific distance from the center.

Step-by-Step Solution

1. Identify the role of the given vertices.

   Let the given vertices be $A(-4,5)$ and $B(3,4)$. The given diagonal is $L \equiv 7x - y + 8 = 0$.

   First, let's check if $A$ or $B$ lie on the diagonal $L$:

   • For $A(-4,5)$: Substitute into $L$: $7(-4) - 5 + 8 = -28 - 5 + 8 = -25$. Since $-25 \neq 0$, $A$ does not lie on $L$.
   • For $B(3,4)$: Substitute into $L$: $7(3) - 4 + 8 = 21 - 4 + 8 = 25$. Since $25 \neq 0$, $B$ does not lie on $L$.

   Since neither $A$ nor $B$ lies on the diagonal $L$, they must be the two vertices that form the other diagonal of the square.

2. Find the center of the square.

   The diagonals of a square bisect each other at the center. Since $A$ and $B$ are opposite vertices, their midpoint is the center of the square.

   Let $M$ be the midpoint of $AB$.

   $M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = \left(\frac{-4 + 3}{2}, \frac{5 + 4}{2}\right) = \left(-\frac{1}{2}, \frac{9}{2}\right)$.

   So, the center of the square is $M\left(-\frac{1}{2}, \frac{9}{2}\right)$.

   Tip

   It's a good sanity check to verify that the center $M$ actually lies on the given diagonal $L$. If it didn't, our initial assumption that $A$ and $B$ are opposite vertices would be wrong, or there's an error in calculation.

   Substitute $M\left(-\frac{1}{2}, \frac{9}{2}\right)$ into $L \equiv 7x - y + 8 = 0$:

   $7\left(-\frac{1}{2}\right) - \frac{9}{2} + 8 = -\frac{7}{2} - \frac{9}{2} + \frac{16}{2} = \frac{-7 - 9 + 16}{2} = \frac{0}{2} = 0$.

   This confirms that $M$ lies on $L$, so our interpretation of $A$ and $B$ as opposite vertices is correct.

3. Calculate the length of the diagonal $AB$.

   The distance formula for $AB$:

   $d_{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$

   $d_{AB} = \sqrt{(3 - (-4))^2 + (4 - 5)^2}$

   $d_{AB} = \sqrt{(7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$.

   In a square, both diagonals have the same length. So, the length of the diagonal $L$ is also $5\sqrt{2}$.

4. Determine the distance from the center to any vertex.

   The center $M$ is the midpoint of both diagonals. Therefore, the distance from the center to any vertex is half the length of a diagonal.

   Distance $MV = \frac{d_{AB}}{2} = \frac{5\sqrt{2}}{2}$.

5. Find the coordinates of the vertices on diagonal $L$.

   Let the two unknown vertices on $L$ be $C(x,y)$ and $D(x',y')$.

   These vertices must satisfy two conditions:

   • They lie on the line $L \equiv 7x - y + 8 = 0$. This means $y = 7x + 8$.
   • Their distance from the center $M\left(-\frac{1}{2}, \frac{9}{2}\right)$ is $\frac{5\sqrt{2}}{2}$.

   Using the distance formula for $MC$:

   $MC^2 = \left(\frac{5\sqrt{2}}{2}\right)^2 = \frac{25 \times 2}{4} = \frac{50}{4} = \frac{25}{2}$.

   So, $(x - x_M)^2 + (y - y_M)^2 = \frac{25}{2}$. …

The concept here is reflection of a point across a line. When a point is reflected across a line, the line acts as a perpendicular bisector of the segment joining the point and its image. So the image $P'$ is found by moving from $P$ to the line along the perpendicular, then going an equal distance on the other side. Once we have $P'$, the distance to another line is just a direct application of the perpendicular distance formula.

Let’s work through it step by step.

1. Find the foot of the perpendicular from $P(-1,2)$ to the line $x - 2y + 3 = 0$. The foot $M$ is the midpoint of $P$ and its image $P'$. For a line $ax + by + c = 0$, the foot of the perpendicular from $(x_1, y_1)$ is given by:

Here $a = 1$, $b = -2$, $c = 3$, and $(x_1, y_1) = (-1, 2)$.

Compute $ax_1 + by_1 + c = 1(-1) + (-2)(2) + 3 = -1 - 4 + 3 = -2$.

So:

Hence $x + 1 = \frac{2}{5} \Rightarrow x = -\frac{3}{5}$, and $y - 2 = -\frac{4}{5} \Rightarrow y = \frac{6}{5}$.

So the foot $M$ is $\left(-\frac{3}{5}, \frac{6}{5}\right)$.

2. Find the image $P'(a, b)$. Since $M$ is the midpoint of $P$ and $P'$:

Solving: $-1 + a = -\frac{6}{5} \Rightarrow a = -\frac{6}{5} + 1 = -\frac{1}{5}$.

And $2 + b = \frac{12}{5} \Rightarrow b = \frac{12}{5} - 2 = \frac{2}{5}$.

So $P' = \left(-\frac{1}{5}, \frac{2}{5}\right)$. …

The concept here is condition for two circles to touch each other externally. When two circles touch externally, the distance between their centres equals the sum of their radii. Additionally, the centre of the first circle must lie in the 4th quadrant — that means its x-coordinate is positive and y-coordinate is negative. This extra condition will help us pick the correct $k$ from the possible values.

Let’s work through it step by step.

1. Rewrite the first circle in standard form.

   The given equation is $x^2 + y^2 + 2kx + 4y - 4 = 0$.

   Complete the square for $x$ and $y$:

   $x^2 + 2kx = (x + k)^2 - k^2$

   $y^2 + 4y = (y + 2)^2 - 4$

   So the equation becomes:

   $(x + k)^2 - k^2 + (y + 2)^2 - 4 - 4 = 0$

   $(x + k)^2 + (y + 2)^2 = k^2 + 8$

   Hence, centre $C_1 = (-k, -2)$ and radius $r_1 = \sqrt{k^2 + 8}$.

2. Rewrite the second circle in standard form.

   The equation is $x^2 + y^2 + 6x - 2y + 6 = 0$.

   Complete the square:

   $x^2 + 6x = (x + 3)^2 - 9$

   $y^2 - 2y = (y - 1)^2 - 1$

   So: $(x + 3)^2 - 9 + (y - 1)^2 - 1 + 6 = 0$

   $(x + 3)^2 + (y - 1)^2 = 4$

   Hence, centre $C_2 = (-3, 1)$ and radius $r_2 = 2$.

3. Apply the condition for external touching.

   Distance between centres $C_1$ and $C_2$:

   $d = \sqrt{(-k + 3)^2 + (-2 - 1)^2} = \sqrt{(3 - k)^2 + (-3)^2} = \sqrt{(3 - k)^2 + 9}$

   For external touch: $d = r_1 + r_2 = \sqrt{k^2 + 8} + 2$

   So: $\sqrt{(3 - k)^2 + 9} = \sqrt{k^2 + 8} + 2$

4. Solve the equation.

   Square both sides:

   $(3 - k)^2 + 9 = k^2 + 8 + 4\sqrt{k^2 + 8} + 4$

   Expand $(3 - k)^2 = k^2 - 6k + 9$, so LHS becomes $k^2 - 6k + 9 + 9 = k^2 - 6k + 18$

   RHS: $k^2 + 12 + 4\sqrt{k^2 + 8}$

   Cancel $k^2$ from both sides:

   $-6k + 18 = 12 + 4\sqrt{k^2 + 8}$

   $-6k + 6 = 4\sqrt{k^2 + 8}$

   Divide by 2: $-3k + 3 = 2\sqrt{k^2 + 8}$ …