Let's break down the problems from the provided image:

Problem 13:

The points P(-5, 6), Q(-3, 2), and R(3, 2) form a triangle.

1. (a) Show that angle PQR is a right angle.

   • To show that $\angle PQR$ is a right angle, we'll check if the vectors $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ are perpendicular by using the dot product. If their dot product is 0, the angle is 90 degrees (right angle).

   Vectors:

   • $\overrightarrow{PQ} = Q - P = (-3 - (-5), 2 - 6) = (2, -4)$
   • $\overrightarrow{QR} = R - Q = (3 - (-3), 2 - 2) = (6, 0)$

   Dot product: $\overrightarrow{PQ} \cdot \overrightarrow{QR} = 2 \times 6 + (-4) \times 0 = 12$ Since the dot product is not zero, $\angle PQR$ is not a right angle.

Problem 5:

The line $2y - x = 12$ intersects the circle $x^2 + y^2 - 10x - 12y + 36 = 0$ at points A and B.

1. (a) Find the coordinates of the points A and B.

   • We substitute $x$ from the line equation into the equation of the circle. First, solve for $x$ in terms of $y$ from $2y - x = 12$, i.e., $x = 2y - 12$.
   • Then substitute into the circle's equation and solve for $y$. After finding $y$, substitute back to find $x$.

2. (b) Find the equation of the perpendicular bisector of $AB$.

   • The perpendicular bisector passes through the midpoint of $A$ and $B$ and is perpendicular to the line segment $AB$.
   • First, find the midpoint of $A$ and $B$, and then determine the slope of the line $AB$. The slope of the perpendicular bisector will be the negative reciprocal of the slope of $AB$.

3. (c) The perpendicular bisector of $AB$ intersects the circle at points $P$ and $Q$. Find the exact coordinates of $P$ and $Q$.

   • After finding the equation of the perpendicular bisector, we set it equal to the equation of the circle and solve for the intersection points.

Problem 6:

Show that the circles $x^2 + y^2 = 25$ and $x^2 + y^2 - 24x - 18y + 125 = 0$ touch each other.

• To show that two circles touch each other, we need to check if the distance between their centers equals the sum or difference of their radii.

  1. The first circle is in the form $x^2 + y^2 = r^2$, which has center $(0, 0)$ and radius $5$ (since $r^2 = 25$).
  2. For the second circle, rewrite it in standard form by completing the square: $(x - 12)^2 + (y - 9)^2 = 12^2 + 9^2 - 125 = 25$ This circle has center $(12, 9)$ and radius $5$.

• Now, calculate the distance between the centers, which is $\sqrt{(12 - 0)^2 + (9 - 0)^2} = \sqrt{144 + 81} = 15$.

• Since the distance between the centers (15) equals the sum of the radii $5 + 5 = 10$, the circles touch each other externally.

Would you like further clarification or details on any specific problem?

Here are 5 related questions to consider:

1. How can you verify if two lines are perpendicular using their slopes?
2. What is the geometric meaning of the dot product being zero?
3. How do you complete the square to rewrite a circle equation in standard form?
4. What are the conditions for two circles to be tangent to each other?
5. How do you calculate the distance between two points in a plane?

Tip: When working with circles, always try to express the equation in the standard form $(x - h)^2 + (y - k)^2 = r^2$ to easily identify the center and radius.