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2.1 In the diagram below, A, B(-5; 12) and C(t; 0) lie on a circle with centre O at the origin - English General - NSC Technical Mathematics - Question 2 - 2021 - Paper 2
Question 2
2.1 In the diagram below, A, B(-5; 12) and C(t; 0) lie on a circle with centre O at the origin. A line is drawn to intersect the circle at B and C. A is on the x-axi... show full transcript
Worked Solution & Example Answer:2.1 In the diagram below, A, B(-5; 12) and C(t; 0) lie on a circle with centre O at the origin - English General - NSC Technical Mathematics - Question 2 - 2021 - Paper 2
Step 1
The equation of the circle
Answer
To find the equation of the circle, we use the standard form ( (x - h)^2 + (y - k)^2 = r^2 ), where ( (h, k) ) is the center and ( r ) is the radius.
Given that the center O is at (0, 0), we need to find the radius using the coordinates of point B(-5, 12):
[ r^2 = (-5)^2 + (12)^2 = 25 + 144 = 169 ]
Thus, the equation of the circle is:
[ x^2 + y^2 = 169 ]
Step 2
Step 3
The equation of the tangent to the circle at B
Answer
The slope of line AB can be determined using the coordinates of points A(5, 0) and B(-5, 12):
[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 0}{-5 - 5} = \frac{12}{-10} = -\frac{6}{5} ]
The equation of the tangent line in point-slope form ( y - y_1 = m(x - x_1) ) using the point B(-5, 12):
[ y - 12 = -\frac{6}{5}(x + 5) ]
Simplifying this, we find:
[ y = -\frac{6}{5}x - 6 + 12 ] [ y = -\frac{6}{5}x + 6 ]
Step 4
Draw the graph defined by \( \frac{x^2}{16} + \frac{y^2}{35} = 1 \)
Answer
To sketch the graph of the equation, note that it represents an ellipse. The intercepts can be found by setting x and y to zero:
- X-intercepts: Set ( y = 0 ):
[ \frac{x^2}{16} = 1 \Rightarrow x^2 = 16 \Rightarrow x = \pm4 ]
- Y-intercepts: Set ( x = 0 ):
[ \frac{y^2}{35} = 1 \Rightarrow y^2 = 35 \Rightarrow y = \pm\sqrt{35} \approx \pm5.92 ]
Now plot the ellipse with intercepts at (4, 0), (-4, 0), (0, \sqrt{35}), and (0, -\sqrt{35}), clearly marking all intercepts on the axes.