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In die diagram is O die middelpunt van die sirkel met vergelyking $x^{2}+y^{2}=20$ - NSC Mathematics - Question 4 - 2023 - Paper 2

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In die diagram is O die middelpunt van die sirkel met vergelyking $x^{2}+y^{2}=20$. G($t; 0$) is die middelpunt van die groter sirkel. Die Gemeenskaplike raaklyn ... show full transcript

Worked Solution & Example Answer:In die diagram is O die middelpunt van die sirkel met vergelyking $x^{2}+y^{2}=20$ - NSC Mathematics - Question 4 - 2023 - Paper 2

Step 1

4.1 Dit word gegee dat D($P_{f}=-2$) op die kleiner sirkel lê. Toon dat $p=4$.

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Answer

Given the equation of the smaller circle, we substitute the point D(Pf=2P_{f}=-2):

x2+(2)2=20x^{2} + (-2)^{2} = 20 p2+4=20p^{2} + 4 = 20 p2=16p^{2} = 16 Thus, p=4p = 4.

Step 2

4.2 E($6; 2$) is die middelpunt van DF. Bepaal die koördinate van F.

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Answer

To find the coordinates of F, we can use the midpoint formula. If E(6,26, 2) is the midpoint between D and F, then:

xF=xD+(2(xExD))x_{F} = x_{D} + (2(x_E - x_D)) xD=4,yD=2x_{D} = 4, y_{D} = -2 This gives: xF=4+2(64)=8x_{F} = 4 + 2(6 - 4) = 8 yF=2+2(2(2))=6y_{F} = -2 + 2(2 - (-2)) = 6 Thus, F is at (8, 6).

Step 3

4.3 Bepaal die vergelyking van die gemeenskaplike raaklyn, DF, in die vorm $y=mx+c$.

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First, we find the slope of line DF (denoted as m):

m=yFyDxFxD=6(2)84=84=2m = \frac{y_{F} - y_{D}}{x_{F} - x_{D}} = \frac{6 - (-2)}{8 - 4} = \frac{8}{4} = 2

Next, we use point-slope form to find the equation of line DF: yyD=m(xxD)y - y_{D} = m(x - x_{D}) Substituting (4, -2): y(2)=2(x4)y - (-2) = 2(x - 4) y+2=2x8y + 2 = 2x - 8 Thus, the equation in slope-intercept form is: y=2x10y = 2x - 10.

Step 4

4.4 Bereken die waarde van $t$. Toon ALLE berekeninge.

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Answer

The relation between the slopes of the lines requires us to find one specific for DF, and the perpendicular slope for the radius at point D. Using the point D(4, -2) and the coordinates of F (8, 6): mDF=6(2)84=2m_{DF} = \frac{6 - (-2)}{8 - 4} = 2

And for the perpendicular GD: mOD=1mDF=12m_{OD} = -\frac{1}{m_{DF}} = -\frac{1}{2}

Substituting in point-slope form: y(2)=12(x4)y - (-2) = -\frac{1}{2}(x - 4) From the calculations, we find that t=20t = 20.

Step 5

4.5 Bepaal die vergelyking van die groter sirkel in die vorm $ax^{2}+by^{2}+cx+dy+e=0$.

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Answer

For the larger circle's equation: x2+y2=180x^{2} + y^{2} = 180 We can rearrange this to the standard form: x2+y2180=0x^{2} + y^{2} - 180 = 0 Thus: a=1,b=1,c=0,d=0,e=180a = 1, b = 1, c = 0, d = 0, e = -180.

Step 6

4.6 Die kleiner sirkel moet $k$ eenhede langs die x-as geatrekker word sodat dit die groter sirkel inwardig raak. Bereken die moontlike waardes van $k$.

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Answer

To find kk, we analyze the horizontal shift as follows: Using the radius of the smaller circle: r=20=25r = \sqrt{20} = 2\sqrt{5} For the larger circle: R=180=65R = \sqrt{180} = 6\sqrt{5} Setting the distance for inward tangency, we find two cases for kk: k=Rr=6525=45k=2(Rr)=2(6525)=85k = R - r = 6\sqrt{5} - 2\sqrt{5} = 4\sqrt{5} \\ k = 2(R - r) = 2*(6\sqrt{5} - 2\sqrt{5}) = 8\sqrt{5} Thus, possible values are: 11.06 and 28.9411.06\text{ and } 28.94.

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