Equation of a Circle

Edexcel IAL WMA12/01/P2 /June/Oct 2020/Q4 (Equation of Circles)

The points P and Q have coordinates (–11, 6) and (–3, 12) respectively.​​ 

Given that PQ is a diameter of the circle C,​​ 

• find the coordinates of the centre of C,​​ 

• find the radius of C.​​ 

(4)

• Hence find an equation of C.​​ 

(2)

• Find an equation of the tangent to C at the point Q giving your answer in the form ax + by + c = 0 where a, b and c are integers to be found.​​ 

(3)

SOLUTION​​ 

a-​​ 

i- The centre can be found by using the midpoint formula as it lies at the centre of QP.​​ 

M.Pt=x1+x22,y1+y22

Center -3+-112   ,   12+62

Center -7 , 9

ii- To find the radius, half the distance PQ.

d=PQ=x2-x12+y2-y12

PQ=-11--32+6-122

r=12 PQ=12-11--32+6-122

r=12-82+-62

r= 12100

r=102

r=5 unit  

b- Plugging the value of radius and centre into the standard form of equation.​​ 

x-a2+y-b2=r2

x--72+y-92=52

x--72+y-92=25

x+72+y-92=  25

c-​​ To find the equation of the tangent of the circle, we will first find the gradient of normal or​​ mOQ​​ to get the gradient of tangent and then use the point-slope formula.​​ 

So, first finding the gradient of normal or​​ mOQ, where O (-7, 9) and Q (-3, 12).

mOQ=y2-y1x2-x1

=9-12-7--3

=-3-4

mOQ=34

Since tangent and normal are perpendicular to each other.

mN×mT=-1

mOQ×mT=-1

mT= -43

Now, using the point-slope formula. Substituting the gradient ​​ and point Q as the tangent passes through it.

y-y1=m(x-x1)

y-y1=mT(x-x1)

y-12= -43 x+3

3y-36= -4x-12

4x+3y-24=0