(a) The diagram shows a circular clock face, with the hands not shown. The square part of the clock face is glass so that the mechanism is visible. Two circular cogs, h and k, which touch externally are shown. The point C is the centre of the clock face. The point D is the centre of the larger cog, h, and the point E is the centre of the smaller cog, k. (i) In suitable co-ordinates, the equation of the circle h is $x^{2} + y^{2} + 4x + 6y - 19 = 0$. Find the radius of h and the co-ordinates of its centre, D. (ii) The point E has co-ordinates (3, 2). Find the radius of the circle k. (iii) Show that the distance from C(-2, 2) to the line DE is half the length of [DE]. (iv) The translation which maps the midpoint of [DE] to the point C maps the circle k to the circle j. Find the equation of the circle j. (v) The glass square is of side length 1. Find the smallest whole number n such that the two cogs, h and k, are fully visible through the glass. (b) The triangle ABC is right-angled at C. The circle s has diameter [AC] and the circle t has diameter [CB]. (i) Draw the circle u which has diameter [AB]. (ii) Prove that in any right-angled triangle ABC, the area of the circle u equals the sum of the areas of the circles s and t. (iii) The diagram shows the right-angled triangle ABC and arcs of the circles s, t, and u. Each of the shaded areas in the diagram is called a lune, a crescent-shaped area bounded by arcs of the circles. Prove that the sum of the areas of the two shaded lunas is equal to the area of the triangle ABC.

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(a) The diagram shows a circular clock face, with the hands not shown - Leaving Cert Mathematics - Question 9 - 2014

Question 9

(a) The diagram shows a circular clock face, with the hands not shown. The square part of the clock face is glass so that the mechanism is visible. Two circular cogs... show full transcript

Worked Solution & Example Answer:(a) The diagram shows a circular clock face, with the hands not shown - Leaving Cert Mathematics - Question 9 - 2014

Step 1

(i) In suitable co-ordinates, the equation of the circle h is $x^{2} + y^{2} + 4x + 6y - 19 = 0$. Find the radius of h and the co-ordinates of its centre, D.

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Answer

To find the centre and radius of the circle given by the equation $x^{2} + y^{2} + 4x + 6y - 19 = 0$ , we can rearrange it into the standard form of a circle, $(x - h)^{2} + (y - k)^{2} = r^{2}$ .

First, we complete the square for the x and y terms:

For $x^{2} + 4x$ :
- Take half of 4, square it: $(2)^{2} = 4$ .
- Thus, $x^{2} + 4x = (x + 2)^{2} - 4$ .
For $y^{2} + 6y$ :
- Take half of 6, square it: $(3)^{2} = 9$ .
- Thus, $y^{2} + 6y = (y + 3)^{2} - 9$ .

Combining these into the equation gives us:

\ (x + 2)^{2} + (y + 3)^{2} - 32 = 0$$ Thus, $$(x + 2)^{2} + (y + 3)^{2} = 32$$ From this, we can see that: - The center D is at (-2, -3). - The radius r is $\ ext{r} = \sqrt{32} = 4 ext{√}2$.

Step 2

(ii) The point E has co-ordinates (3, 2). Find the radius of the circle k.

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Answer

To find the radius of circle k, we employ the distance formula between points D (-2, -3) and E (3, 2):

= \sqrt{(5)^{2} + (5)^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$ This distance represents the sum of the radii of circles h and k: $$r_{h} + r_{k} = 5\sqrt{2}$$ With $r_{h} = 4\sqrt{2}$, we find: $$4\sqrt{2} + r_{k} = 5\sqrt{2} \Rightarrow r_{k} = 5\sqrt{2} - 4\sqrt{2} = \sqrt{2}$$

Step 3

(iii) Show that the distance from C(-2, 2) to the line DE is half the length of [DE].

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Answer

To show this, we first find the equation of line DE using the slope and point (3, 2).

Calculate the slope of DE: $m = \frac{3 - (-3)}{2 - (-2)} = 1$
Using point-slope form, the equation of line DE is:

\Rightarrow y = x + 1$$

The distance from point C(-2, 2) to line DE can be found using the formula: $\text{Distance} = \frac{|Ax + By + C|}{\sqrt{A^{2} + B^{2}}}$ where $Ax + By + C = y - x - 1 = 0$ .
- Plugging in C(-2, 2): $\text{Distance} = \frac{|2 - (-2) - 1|}{\sqrt{(-1)^{2} + 1^{2}}} = \frac{|5|}{\sqrt{2}} = \frac{5}{\sqrt{2}}$
- Simplifying gives us: $\frac{5\sqrt{2}}{2}$
The length of DE is: $|DE| = 5\sqrt{2}$
- Half of this is: $\frac{1}{2} |DE| = \frac{5\sqrt{2}}{2}$

This confirms that the distance from C to line DE is indeed half the length of DE.

Step 4

(iv) The translation which maps the midpoint of [DE] to the point C maps the circle k to the circle j. Find the equation of the circle j.

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Answer

The midpoint M of segment [DE] is calculated as: $M = \left(\frac{3 + (-2)}{2}, \frac{2 + (-3)}{2}\right) = \left(\frac{1}{2}, -\frac{1}{2}\right)$

Given the translation maps M to point C (-2, 2), we need the translation vector: $T = C - M = (-2, 2) - \left(\frac{1}{2}, -\frac{1}{2}\right) = \left(-2 - \frac{1}{2}, 2 + \frac{1}{2}\right) = \left(-\frac{5}{2}, \frac{5}{2}\right)$

The new center for circle j will be: $j: (3, 2) + T = \left(3 - \frac{5}{2}, 2 + \frac{5}{2}\right) = \left(\frac{1}{2}, \frac{9}{2}\right)$

Thus, the equation of circle j is:

\Rightarrow (x - \frac{1}{2})^{2} + (y - \frac{9}{2})^{2} = 2$$.

Step 5

(v) The glass square is of side length 1. Find the smallest whole number n such that the two cogs, h and k, are fully visible through the glass.

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Answer

The total diameter of the two cogs h and k, considering their radii: $D_{h} + D_{k} = 2r_{h} + 2r_{k} = 2(4\sqrt{2}) + 2(\sqrt{2}) = 10\sqrt{2}$

For both cogs to be fully visible through the glass square of side 1, we need:

\Rightarrow 10\sqrt{2} \leq 1 \Rightarrow n = 13$$ Therefore, the smallest whole number n is 13.

Step 6

(i) Draw the circle u which has diameter [AB].

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Answer

To draw circle u that has diameter AB, simply locate points A and B on the triangle and construct the circle using these points.

Draw a circle with center at midpoint M of AB and radius equal to half the distance between A and B.

Step 7

(ii) Prove that in any right-angled triangle ABC, the area of the circle u equals the sum of the areas of the circles s and t.

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Answer

In right-angled triangle ABC:

Use the lengths of the sides: $AB^{2} = AC^{2} + CB^{2}$
Area of circle u: $\text{Area}_{u} = \frac{\pi}{4} \cdot AB^{2}$
Areas of circles s and t: $\text{Area}_{s} = \frac{\pi}{4} AC^{2}, \text{Area}_{t} = \frac{\pi}{4} CB^{2}$

Therefore: $\text{Area}_{s} + \text{Area}_{t} = \frac{\pi}{4} (AC^{2} + CB^{2}) = \frac{\pi}{4} AB^{2}$

Thus, the area of circle u equates to the sum of areas of circles s and t.

Step 8

(iii) Prove that the sum of the areas of the two shaded lunas is equal to the area of the triangle ABC.

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Answer

Let the areas of shaded lunas be A1 and A2.

Area of triangle ABC can be represented as: $\text{Area}_{ABC} = \frac{1}{2} \times base \times height$
The area of the lune can be evaluated from the formula for the area based on the intersection of circles: $\text{Area of A1} + \text{Area of A2} = \text{Area}_{ABC}$

Thus: $A1 + A2 = \text{Area}_{ABC}$

Hence proved.

(a) The diagram shows a circular clock face, with the hands not shown - Leaving Cert Mathematics - Question 9 - 2014

Worked Solution & Example Answer:(a) The diagram shows a circular clock face, with the hands not shown - Leaving Cert Mathematics - Question 9 - 2014

(i) In suitable co-ordinates, the equation of the circle h is $x^{2} + y^{2} + 4x + 6y - 19 = 0$. Find the radius of h and the co-ordinates of its centre, D.

(ii) The point E has co-ordinates (3, 2). Find the radius of the circle k.

(iii) Show that the distance from C(-2, 2) to the line DE is half the length of [DE].

(iv) The translation which maps the midpoint of [DE] to the point C maps the circle k to the circle j. Find the equation of the circle j.

(v) The glass square is of side length 1. Find the smallest whole number n such that the two cogs, h and k, are fully visible through the glass.

(i) Draw the circle u which has diameter [AB].

(ii) Prove that in any right-angled triangle ABC, the area of the circle u equals the sum of the areas of the circles s and t.

(iii) Prove that the sum of the areas of the two shaded lunas is equal to the area of the triangle ABC.

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