In die diagram is P(−3; 4) die middelpunt van die sirkel. V(k; 1) en W is die eindpunte van 'n middellyn. Die sirkel sny die y-as by B en C. BCVW is 'n koordeverhoog. CV is verleng om die x-as by T te sny. Of C = a. 4.1 Die radius van die sirkel is $\sqrt{10}$. Bereken die waarde van $k$ as punt V regs van punt P geleë is. Toon ALLE berekening duidelik. 4.2 Die vergelyking van die sirkel word as $x^{2} + 6x + y^{2} - 8y + 15 = 0$ gegee. Bereken die lengte van BC. 4.3 As $k = -2$, bereken die grotte van: 4.3.1 $\alpha$ 4.3.2 VWB 4.4 'n Nuwe sirkel word verkry wanneer die gegewe sirkel om die lyn $y = 1$ gerflikteer word. Bepaal die: 4.4.1 Koördinate van Q, die middelpunt van die nuwe sirkel 4.4.2 Vergelyking van die nuwe sirkel in die vorm $(x - a)^{2} + (y - b)^{2} = r^{2}$ 4.4.3 Vergelykings van die lyne wat ewewydig aan die y-as gekke is en deur die snypunte van die twee sirkel gaan.

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In die diagram is P(−3; 4) die middelpunt van die sirkel - NSC Mathematics - Question 4 - 2021 - Paper 2

Question 4

In die diagram is P(−3; 4) die middelpunt van die sirkel. V(k; 1) en W is die eindpunte van 'n middellyn. Die sirkel sny die y-as by B en C. BCVW is 'n koordeverhoog... show full transcript

Worked Solution & Example Answer:In die diagram is P(−3; 4) die middelpunt van die sirkel - NSC Mathematics - Question 4 - 2021 - Paper 2

Step 1

4.1 Die radius van die sirkel is $\sqrt{10}$. Bereken die waarde van $k$ as punt V regs van punt P geleë is. Toon ALLE berekening duidelik.

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Answer

To determine the value of $k$ , we start by applying the distance formula between points P and V. We know the coordinates for P are $(-3, 4)$ and for V are $(k, 1)$ :

$PV = \sqrt{(k - (-3))^{2} + (1 - 4)^{2}} = \sqrt{(k + 3)^{2} + (-3)^{2}} = \sqrt{(k + 3)^{2} + 9}$

Setting the distance $PV = \sqrt{10}$ , we solve:

$\sqrt{(k + 3)^{2} + 9} = \sqrt{10}$

Squaring both sides leads to:

$(k + 3)^{2} + 9 = 10$

$(k + 3)^{2} = 1$

Taking the square root gives:

$k + 3 = 1$ or $k + 3 = -1$ .

Thus, $k = -2$ or $k = -4$ . Since we seek the value of k where point V is to the right of point P, we discard $k = -4$ and accept $k = -2$ .

Step 2

4.2 Die vergelyking van die sirkel word as $x^{2} + 6x + y^{2} - 8y + 15 = 0$ gegee. Bereken die lengte van BC.

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Answer

To find the length of BC, we first convert the equation into standard form.

Completing the square gives:

$x^{2} + 6x + y^{2} - 8y + 15 = 0$

For $x$ : $x^{2} + 6x = (x + 3)^{2} - 9$

For $y$ : $y^{2} - 8y = (y - 4)^{2} - 16$

Substituting back, we have:

$(x + 3)^{2} + (y - 4)^{2} - 10 = 0$

Thus, the center is at $(-3, 4)$ and diameter = $2\sqrt{10}$ , making radius = $\sqrt{10}$ . The y-intercepts are determined from setting $x = 0$ :

$0 + 6(0) + y^{2} - 8y + 15 = 0$

This simplifies to yield the y-coordinates for points B and C, then applying the distance formula gives the length BC = 2 units.

Step 3

4.3.1 As $k = -2$, bereken die grotte van: $\alpha$

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Answer

Given $k = -2$ , we derive the slope of line BC using:

$m_{BC} = \frac{y_{C} - y_{B}}{x_{C} - x_{B}}$

Substituting values from our previous results leads us to:

$m_{BC} = \frac{3 - 5}{0 - (-4)} = \frac{-2}{4} = -\frac{1}{2}$

Then we find $\alpha$ using:

$tan(\alpha) = \left|\frac{1 + m_{BC}}{1 - m_{BC}}\right|$

Where $tan(\alpha)$ simplifies to $1$ , leading us to conclude that:

$\alpha = 45^{\circ}$ .

Step 4

4.3.2 VWB

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The value of VWB can be deduced through the properties of cyclic quadrilaterals, relationship between opposite angles, yielding:

$VWB = 135^{\circ}$ .

Step 5

4.4.1 Koördinate van Q, die middelpunt van die nuwe sirkel

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To find the coordinates of Q, the new center, we reflect the original center P at the line $y = 1$ . The transformation yields:

If P is at $(−3, 4)$ , then:

$Q = (−3, 1 + (1 − 4)) = (−3, −2)$

Step 6

4.4.2 Vergelyking van die nuwe sirkel in die vorm $(x - a)^{2} + (y - b)^{2} = r^{2}$

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The new circle has center Q at $(−3, −2)$ and maintaining radius of $\sqrt{10}$ . Thus, the equation is:

$(x + 3)^{2} + (y + 2)^{2} = 10$ .

Step 7

4.4.3 Vergelykings van die lyne wat ewewydig aan die y-as gekke is en deur die snypunte van die twee sirkel gaan.

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Answer

Finally, for the equations of lines parallel to the y-axis through intersection points: Using the circle equations derived, intersection points found earlier provide vertical lines at:

$x = -4 \text{ and } x = 0$ .

In die diagram is P(−3; 4) die middelpunt van die sirkel - NSC Mathematics - Question 4 - 2021 - Paper 2

Worked Solution & Example Answer:In die diagram is P(−3; 4) die middelpunt van die sirkel - NSC Mathematics - Question 4 - 2021 - Paper 2

4.1 Die radius van die sirkel is $\sqrt{10}$. Bereken die waarde van $k$ as punt V regs van punt P geleë is. Toon ALLE berekening duidelik.

4.2 Die vergelyking van die sirkel word as $x^{2} + 6x + y^{2} - 8y + 15 = 0$ gegee. Bereken die lengte van BC.

4.3.1 As $k = -2$, bereken die grotte van: $\alpha$

4.3.2 VWB

4.4.1 Koördinate van Q, die middelpunt van die nuwe sirkel

4.4.2 Vergelyking van die nuwe sirkel in die vorm $(x - a)^{2} + (y - b)^{2} = r^{2}$

4.4.3 Vergelykings van die lyne wat ewewydig aan die y-as gekke is en deur die snypunte van die twee sirkel gaan.

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