Equation of a Circle

Edexcel IAL WMA12/01/P2 /Oct 2022/Q9 (Equations of Circles, Discriminant, Integration)

In this question you must show detailed reasoning.​​ 

Solutions relying entirely on calculator technology are not acceptable.

Figure 3 shows​​ 

• the curve C1​​ with equation​​ y = x3 – 5x2 + 3x + 14 

• the circle C2​​ with centre T

The point T is the minimum turning point of C1​​ 

Using Figure 3 and calculus,​​ 

• find the coordinates of T​​ 

(3)

The curve C1​​ intersects the circle C2​​ at the point A with x coordinate 2​​ 

• Find an equation of the circle C2​​ 

(3)

The line l shown in Figure 3, is the tangent to circle C2​​ at A​​ 

• Show that an equation of l is​​ 

y=13x+223

(3)

The region R, shown shaded in Figure 3, is bounded by C1, l and the y‑axis.​​ 

• Find the exact area of R.​​ 

(3)

SOLUTION

a- To find the minimum turning point, let us first diffrentiate the given equation of the curve. Then by looking at the values of​​ x,​​ which ever point is farthest away from the origin is the minimum turning point. It is so as it clear from the figure above in the question that there comes first a​​ maximum turning point and then comes the minimum turning point of the curve.  ​​​​ 

dydx=3x2-10x+3

3x2-10x+3=0

x-33x-1=0

x-3=0          3x-1=0

x=3         x=13

We will consider x cordinate of minimum turning point as 3 because it is greatest value between the two values of x.​​ 

Hence, putting the value of​​ x​​ as​​ 3​​ in the equation of curve to get y-cordinate of the respective point turnming point.​​ 

y=33-532+33+14=5

T3, 5

b- The equation of a circle is given as ​​ 

For deriving the equation of a circle, we need the centre point and the radius of the circle. We have already found the centre point of circle (a,b) in part (a) as​​ T3, 5. For radius, we will find the distance/length AT, where A has an x cordinate as 2. So for y-cordinate, put the value of x in equation of curve C1.

y=2x3-522+32+14

y=8

So, the coordinate of A are (2,8) and T is (3, 5).​​ 

Now, finding the distance AT.

r=x2-x12+y2-y12

r=AT=3-22-8-52=12+32=10

Now, putting the centre point​​ 3, 5​​ and radius (10) in the equation of a circle.​​ 

x-a2+y-b2=r2

x-32+y-52=102

x-32+y-52=10

c- ​​ Use the point slope formula to derive the equation of line.​​ 

To find the equation of line l, consider the given information in the question that line l is tangent to circle. This reminds us​​ of the circle property that tangent (which in our case is line l) is perpendicular to radius of circle:​​ l ⊥AT.

First, let us find the gradient of the radius or AT.​​ 

MAT=8-52-3=3-1=-3

Since it is perpendicular so the gradient of line​​ l​​ can be found as

Ml x MAT=-1

Ml x-3=-1

Ml=13

Now, using the point slope formula to get the equation of a circle. We have a point through which the line l passes point​​ A 2, 8.​​ 

y-y1=m(x-x1)

y-8=13(x-2)

y-8=13x-23

y=13x-23+8

y=13x+223

d- ​​ To find the area of shaded region, integration of the curve.​​ 

With the help of integration of the curve within the limits 0 and 2, we will get the entire area bounded between the curve and x-axis. Now, to extract the area of the shaded region as shown in the question, we have to subtract the area of trapezium from the answer/area we get from the integration.​​ 

Area of R=(Area bounded between curve and x-axis)-(Area of Trapezium)

Area of R=∫02x3-5x2+3x+14dx-12×sum of parallel sides×height

Area of R=x44-5x33+3x22 +14x 02-12 ×223+8×2

Area of R=164-583+342+142-463

Area of R=283 square