Equation of a Circle

Edexcel IAL WMA12/01/P2 /Jan 2021/Q9 (Equation of Circles Equation of Normals)

A circle C has equation​​ 

x - k2 +  y - 2k2 = k + 7

where k is a positive constant.​​ 

• Write down, in terms of k,​​ 

• the coordinates of the centre of C,​​ 

• the radius of C.​​ 

(2)​​ 

Given that the point​​ P(2,3)​​ lies on C​​ 

• ​​ 

• show that​​ 5k2 - 17k + 6 = 0​​ 

• hence find the possible values of k.​​ 

(3)​​ 

The tangent to the circle at P intersects the x-axis at point T.​​ 

Given that k < 2​​ 

• calculate the exact area of triangle OPT.​​ 

(5)

SOLUTION​​ 

a-​​ 

Compare with the standard equation, where (a, b) is centre point and r is the radius.

i- ​​ The equation is given​​ 

x – k2 +  y – 2k2 = k + 7

 center k , 2k

ii-​​ 

r=k+7  

b- i- ​​ Since point P (2, 3) lies in C so it must satisfy the equation.

x - k2 +  y - 2k2 = k + 7

2-k2+3-2k2=k+7

4-4k+k2+9-12k+4k2=k+7

5k2-17k+6=0

ii- Solving quadratic equatoion found above to find the value of k.

5k-2k-3=0

5k-2=0         k-3=0

k=25         k=3

c-​​ 

x - k2 +  y - 2k2 = k + 7

x-252+y-452=25+7

x-252+y-452=375

Hence, centre and radius of a circle are

C25,45 & r=375  

Now, to find the​​ Equation of tangent.

mN=3-0.82-0.9=2.21.6=118

Now,​​ 

mN×mT=-1

MT= -811

Using point-slope formula.

y-y1=m x-x1

y-3=-811 x-2

11y-33= -8x+16 

8x+11y-49=0