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In the diagram, N is the centre of the circle - NSC Mathematics - Question 4 - 2017 - Paper 2

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In the diagram, N is the centre of the circle. M(-3; -2) and P(1; 4) are points on the circle. MNP is the diameter of the circle. Tangents drawn to circle N from poi... show full transcript

Worked Solution & Example Answer:In the diagram, N is the centre of the circle - NSC Mathematics - Question 4 - 2017 - Paper 2

Step 1

Determine the coordinates of N.

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Answer

To find the coordinates of N, which is the center of the circle formed by the diameter MN, we calculate the midpoint of M and P. The coordinates of M are (-3, -2) and P are (1, 4).

Using the midpoint formula: N=(x1+x22,y1+y22)=(3+12,2+42)=(1,1)N = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-3 + 1}{2}, \frac{-2 + 4}{2} \right) = \left( -1, 1 \right)

Thus, the coordinates of N are (-1, 1).

Step 2

Determine the equation of the circle in the form $(x-a)^2 + (y-b)^2 = r^2$.

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The center of the circle is N(-1, 1) and we need to find the radius. The radius can be determined by the distance from N to either point M or P. Let's calculate the distance from N to M:

Using the distance formula: r=(x2x1)2+(y2y1)2=(3(1))2+(21)2=(2)2+(3)2=4+9=13r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-3 - (-1))^2 + (-2 - 1)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}

Thus, the equation of the circle can be written as: (x+1)2+(y1)2=13.(x + 1)^2 + (y - 1)^2 = 13.

Step 3

Determine the equation of the tangent RM in the form $y = mx + c$.

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Answer

The point M lies on the circle N, and the tangent RM can be found using the point-slope form. The slope of line RM can be computed by taking the differences in y-coordinates and x-coordinates from points R and M.

Let the coordinates of R be (x_R, y_R). The slope mm of RM is given as: mRM=yR(2)xR(3)=yR+2xR+3m_{RM} = \frac{y_R - (-2)}{x_R - (-3)} = \frac{y_R + 2}{x_R + 3}

Using the property of tangents to a circle where the tangent is perpendicular to the radius at the point of tangency (M), we find: mRMmNM=1m_{RM} \cdot m_{NM} = -1

Thus, compute the slope of NM and find R's coordinates accordingly.

Step 4

If it is given that the line joining S to M is perpendicular to the x-axis, determine the coordinates of S.

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Answer

Since the line joining S to M is perpendicular to the x-axis, this means that S must have the same x-coordinate as M. The coordinates of M are (-3, -2) therefore, S will have coordinates of the form (-3, y_S).

To find y_S, we will use the equation of the circle or the property of tangents. Given that the radius NM is vertical, the y-coordinate for S must provide a location where the tangent is drawn vertically. Thus, we use the calculated radius to find this exact point.

Step 5

Determine the coordinates of R, the common external point from which both tangents to the circle are drawn.

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Answer

To find the coordinates of R, we equate the lengths of the tangents from R to points M and S. The distances from R(x_R, y_R) to M(-3, -2) and from R to S(-3, y_S) must equal:

(xR+3)2+(yR+2)2=r2(x_R + 3)^2 + (y_R + 2)^2 = r^2

Given that these tangents are equal by nature, we can solve the equations simultaneously to obtain the coordinates of R.

Step 6

Calculate the area of RSNM.

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Answer

To calculate the area of the quadrilateral RSNM, we can split it into two triangles, RSN and RMN. The formula for the area of triangles can be used:

Area=12×base×heightArea = \frac{1}{2} \times base \times height

Calculate area for both triangles separately, taking the bases as the distances between the relevant points and the height as determined earlier. Sum the areas to find the total area of RSNM.

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