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Question 10
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(a) The equation of a circle is x2 + y2 – 4x + 2y + c = o, where c is a constant. If the radius of the circle is 2√3, find the value of c.
(b) T is the tangent to the curve y = x2 + 6x – 4 at (1,3) and N is the normal to the curve y = x2 - 6x + 18 at (4,10). Find the coordinates of the point of intersection of T and N.
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In the part (a), two approaches were open to candidates. Completing the squares approach where the equation could be expressed as (x - 2)2 + (y + 1)2 = 5 – c. Comparing this with the general equation (x - 2)2 + (y + 1 )2 = r2, 5 - c = r2 = (2√3)2 = 12,
Therefore c = 5 – 12 = -7. The second approach is by using the relation r2 = f2 + g2 -c where g = 2, f = -1.
In the part (b), candidates’ performance was not satisfactory. They were expected to find the slope of the tangent T by determining the differential dy of y = x2 + 6x – 4 at point dx
(1,3) which is 8 and use it to obtain the equation of the tangent y = 8x – 5. Also, the differential of y = x2- 6x + 8 at the point (4,10) is 2. Hence the slope of the normal to the curve at this point = -1/2 from where the equation of the normal is gotten as 2y + x - 24 = 0. Solving these two equations simultaneously, we obtain the required solution (x, y) = (2, 11).
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