In the diagram, P(-3; 4) is the centre of the circle. V(k; 1) and W are the endpoints of a diameter. The circle intersects the y-axis at B and C. BCVW is a cyclic quadrilateral. CV is produced to intersect the x-axis at T. O is the origin. 4.1 The radius of the circle is \( \sqrt{10} \). Calculate the value of k if point V is to the right of point P. Clearly show ALL calculations. 4.2 The equation of the circle is given as \( x^2 + 6x + y^2 - 8y + 15 = 0 \). Calculate the length of BC. 4.3 If k = -2, calculate the size of: 4.3.1 \( \alpha \) 4.3.2 VW 4.4 A new circle is obtained when the given circle is reflected about the line \( y = 1 \). Determine: 4.4.1 Coordinates of Q, the centre of the new circle. 4.4.2 Equation of the new circle in the form \( (x - a)^2 + (y - b)^2 = r^2 \) 4.4.3 Equations of the lines parallel to the y-axis and passing through the points of intersection of the two circles.

NSC Mathematics Euclidean Geometry: In the diagram, P(-3; 4) is the

In the diagram, P(-3; 4) is the centre of the circle - NSC Mathematics - Question 4 - 2021 - Paper 2

Question 4

In the diagram, P(-3; 4) is the centre of the circle. V(k; 1) and W are the endpoints of a diameter. The circle intersects the y-axis at B and C. BCVW is a cyclic qu... show full transcript

Worked Solution & Example Answer:In the diagram, P(-3; 4) is the centre of the circle - NSC Mathematics - Question 4 - 2021 - Paper 2

Step 1

4.1 The value of k if point V is to the right of point P

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Answer

To find the value of k, we will use the distance formula between the points P(-3, 4) and V(k, 1) which is equal to the radius ( r = \sqrt{10} ).

Using the distance formula:

PV = \sqrt{(k - (-3))^2 + (1 - 4)^2}

Substituting the values:

PV = \sqrt{(k + 3)^2 + (-3)^2} = \sqrt{10}

Squaring both sides gives:

(k + 3)^2 + 9 = 10

This simplifies to:

(k + 3)^2 = 1

Thus:

k + 3 = 1 \quad \Rightarrow \quad k = -2

Or:

k + 3 = -1 \quad \Rightarrow \quad k = -4

Since V must be to the right of P, choose ( k = -2 ).

Step 2

4.2 Calculate the length of BC

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Answer

To find the length of BC, we need to identify the y-intercepts of the circle equation:

x^2 + 6x + y^2 - 8y + 15 = 0

Setting ( x = 0 ):

0^2 + 6(0) + y^2 - 8y + 15 = 0 \Rightarrow y^2 - 8y + 15 = 0

Factoring:

(y - 5)(y - 3) = 0

Thus:

y = 5 \; \text{and} \; y = 3

The points B and C are at (0, 5) and (0, 3) respectively. The length of BC is:

BC = |5 - 3| = 2

Step 3

4.3.1 \( \alpha \)

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Answer

With ( k = -2 ), point V becomes V(-2, 1). To find ( \alpha ): Using the gradient formula:

m_{CV} = \frac{1 - 4}{-2 - (-3)} = \frac{-3}{1} = -3

Then,

\alpha = \tan^{-1}(-3) \Rightarrow \alpha \approx 135^{\circ}

Step 4

4.3.2 VW

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Answer

Since ( \alpha ) = 135° and VW is to the left of CV, then:

VW = 45^{\circ} \Rightarrow \text{the angle is also } 45^{\circ}

Step 5

4.4.1 Coordinates of Q, the centre of the new circle

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Answer

The original centre Q is at (-3, 4). Reflecting about the line ( y = 1 ), the new y-coordinate becomes:

y_{new} = 1 - (4 - 1) = -1

Thus, Q's new coordinates:

Q(-3, -1)

Step 6

4.4.2 Equation of the new circle

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Answer

The radius remains the same as it is not affected by reflection: Radius = ( r = \sqrt{10} ). Using the center Q(-3, -1):

(x + 3)^2 + (y + 1)^2 = 10

Step 7

4.4.3 Equations of the lines parallel to the y-axis

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Answer

The lines parallel to the y-axis passing through the intersection points can be written as:

x = -2 \quad \text{and} \quad x = -4

In the diagram, P(-3; 4) is the centre of the circle - NSC Mathematics - Question 4 - 2021 - Paper 2

Worked Solution & Example Answer:In the diagram, P(-3; 4) is the centre of the circle - NSC Mathematics - Question 4 - 2021 - Paper 2

4.1 The value of k if point V is to the right of point P

4.2 Calculate the length of BC

4.3.1 \( \alpha \)

4.3.2 VW

4.4.1 Coordinates of Q, the centre of the new circle

4.4.2 Equation of the new circle

4.4.3 Equations of the lines parallel to the y-axis

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