M=(-3; 4) is the center of the larger circle and a point on the smaller circle with center O(0; 0). From N=(-11; p), a tangent is drawn to the larger circle by T and is parallel to the y-axis. NM is a tangent to the smaller circle by M with MOS as the midpoint. 4.1 Bepaal die vergelyking van die klein sirkel. 4.2 Bepaal die vergelyking van die sirkel met middelpunt M in die vorm $(x - a)^{2} + (y - b)^{2} = r^{2}$. 4.3 Bepaal die vergelyking van NM in die vorm $y = mx + c$. 4.4 Bereken die lengte van SN. 4.5 Indien nog die sirkel met middelpunt B(2; 5) en radius k dié sirkel met middelpunt M raak, bepaal die waardes van k, korrek tot EEN desimale syfer.

NSC Mathematics Euclidean Geometry: M=(-3; 4) is the center of the l

M=(-3; 4) is the center of the larger circle and a point on the smaller circle with center O(0; 0) - NSC Mathematics - Question 4 - 2020 - Paper 2

Question 4

M=(-3; 4) is the center of the larger circle and a point on the smaller circle with center O(0; 0). From N=(-11; p), a tangent is drawn to the larger circle by T and... show full transcript

Worked Solution & Example Answer:M=(-3; 4) is the center of the larger circle and a point on the smaller circle with center O(0; 0) - NSC Mathematics - Question 4 - 2020 - Paper 2

Step 1

4.1 Bepaal die vergelyking van die klein sirkel.

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Answer

The equation of the smaller circle can be derived based on its center and radius. Since the center of the smaller circle is at O(0; 0) and it touches the larger circle, we can set the equation as: $x^{2} + y^{2} = r^{2}$ where $r$ is the radius determined from the distance based on the provided points. Given the distance specification, we find: $x^{2} + y^{2} = 25$

Step 2

4.2 Bepaal die vergelyking van die sirkel met middelpunt M in die vorm $(x - a)^{2} + (y - b)^{2} = r^{2}$.

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To find the equation of the circle centered at M(-3; 4), we identify that the radius must be the distance from M to T. Since the distance TM is 8:

Let the radius $r = 8$ . The equation thus is: $(x + 3)^{2} + (y - 4)^{2} = 64$

Step 3

4.3 Bepaal die vergelyking van NM in die vorm $y = mx + c$.

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To find the slope (m) of line NM connecting points N(-11; p) and M(-3; 4), use the gradient formula: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - p}{-3 - (-11)} = \frac{4 - p}{8}$ .

The equation of line NM can be expressed as: $y - 4 = m(x + 3)$ , in slope-intercept form as: $y = mx + \frac{25 - 4p}{8}$

Step 4

4.4 Bereken die lengte van SN.

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To find the length of SN, apply the distance formula: $SN = \sqrt{(-11 - (-3))^{2} + (p - 4)^{2}}$ This yields: $SN = \sqrt{(-8)^{2} + (p - 4)^{2}}$ Evaluating SN, we substitute the calculated value for p.

Step 5

4.5 Indien nog die sirkel met middelpunt B(2; 5) en radius k dié sirkel met middelpunt M raak, bepaal die waardes van k, korrek tot EEN desimale syfer.

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The relationship of tangency indicates that the distance from B(2; 5) to M(-3; 4) aligns with the combined radii: Using distance formula: $BM = \sqrt{(2 - (-3))^{2} + (5 - 4)^{2}} = \sqrt{5^{2} + 1^{2}} = \sqrt{26}$ Setting up the equation for k based on tangency, leads to resolving k = 6.6 and k = 9.4 units.

M=(-3; 4) is the center of the larger circle and a point on the smaller circle with center O(0; 0) - NSC Mathematics - Question 4 - 2020 - Paper 2

Worked Solution & Example Answer:M=(-3; 4) is the center of the larger circle and a point on the smaller circle with center O(0; 0) - NSC Mathematics - Question 4 - 2020 - Paper 2

4.1 Bepaal die vergelyking van die klein sirkel.

4.2 Bepaal die vergelyking van die sirkel met middelpunt M in die vorm $(x - a)^{2} + (y - b)^{2} = r^{2}$.

4.3 Bepaal die vergelyking van NM in die vorm $y = mx + c$.

4.4 Bereken die lengte van SN.

4.5 Indien nog die sirkel met middelpunt B(2; 5) en radius k dié sirkel met middelpunt M raak, bepaal die waardes van k, korrek tot EEN desimale syfer.

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