Equation of a Circle

Edexcel IAL WMA12/01/P2 /Jan 2022/Q6 (Equation of Circles)

The points P(23, 14), Q(15, –30) and R(–7, –26) lie on the circle C, as shown in Figure 1.​​ 

• Show that angle PQR = 90°​​ 

(2)​​ 

• Hence, or otherwise, find​​ 

• the centre of C,​​ 

• the radius of C.​​ 

(3)​​ 

Given that the point S lies on C such that the distance QS is greatest,​​ 

 ​​ ​​ ​​ ​​ ​​​​ (c) find an equation of the tangent to C at S, giving your answer in the form ax + by + c = 0, where a, b, and c are integers to be found.​​ 

(3)

SOLUTION​​ 

a-​​ 

If​​ angle​​ PQR=90°, then the lones PQ and QR must be perpendicular meaning thereby that​​ mPQx mQR=-1. ​​ Therefore, we will find​​ mPQ​​ &​​ mQR​​ and check whether their product is equal to -1. ​​ 

mPQ= 14--3023-15=448÷4=112

mQR=--30--2615--7=-422÷ = -211

mPQx mQR =112  x-211= -1  

∴PQ ⊥QR

Hence,​​ PQ^R=90o​​ is shown.​​ 

b-​​ 

PR is the diameter of the circle. So the find the centre of the circle, we should find the midpoint of PR.​​ 

M.Pt=x1+x22,y1+y22

=-7- 232,-26+142

M8, -6

Thus, the centre is 8, -6.​​ 

(For radius, will find the distance between the point centre and anypoint lying on the circumference. The alternate method is that one may find the distance between point P and R, which is the diameter and divide the answer by 2 to get radius. ​​ We will be considering the distance between P (23, 14) and R (-7, -26). The metter method is to find the distance between P and R because in case if the student has found the incoirrect cordinates of the centre and considers that to find the radius, the answer for part b would also get incorrect. So it better to use the points already given in the question.)

r=12 d

d=PR=x2-x12+y2-y12

d= -7-232+-26-142  

d=302+402

=280  

d=50

r=1250

r=25

c-

​​ 

Since QS is the greatest distance then the line QS must have pass through the centre point of the circle,​​ 

Using mid point formula.

M.Pt=x1+x22,y1+y22

(8, -6)=x+152,y-62

Equating the both x and y cordinates separately.​​ 

15+x2=8           -30+y2=-6

15+x=16         -30+y= -12 

x=1                       y=18

Thus, the coordinates of S are​​ 1, 18.

Now, the tangent to C at S will be perpendicular to QS. So finding​​ mQS​​ and then will find the​​ mT.

Finding​​ mQS​​ when Q(15, -30) and S(1,18).

mQS=18--61-8=24-7

Now,​​ 

mQS×mT=-1

24-7=×mT=-1

mT=724 

Now, using point-slope formula.​​ 

y-y1=mx-x1

y-18=724x-1

24 y-432=7x-7

0=7x-24y-425

7x-24y+425=0