Equation of a Circle

Edexcel IAL WMA12/01/P2 /Jan 2023/Q6 (Equations of Circles, Normals)

The circle C has equation​​ 

x2 + y2 + 8x – 4y = 0 

• Find​​ 

• the coordinates of the centre of C,​​ 

• the exact radius of C.​​ 

(3)

The point P lies on C.​​ 

Given that the tangent to C at P has equation​​ x + 2y + 10 = 0​​ 

• find the coordinates of P​​ 

(4)

• Find the equation of the normal to C at P, giving your answer in the form y = mx + c where m and c are integers to be found.​​ 

(3)

SOLUTION​​ 

a- i-​​ 

x2+8x+y2-4y=0

Completing the square for​​ x​​ terms and​​ y​​ terms.​​ 

So for​​ x​​ terms,​​ 

x2+8x=x+822-822=x+42-16

And, for​​ y​​ terms,​​ 

y2-4y=y-422-422=y-22-4

On writing both terms compoleted square together in the parent equation gives,​​ 

x+42-16+y-22-4=0

x+42+y-22-20=0 

x+42+y-22=20

Deducing the centre of the circle from the equation of the circle.​​ 

On comparing​​ x+42+y-22=20​​ with​​ x-a2+y-b2=r2, the centre of the circle comes out to be:

x-a2=x+42

a=-4

And​​ 

y-b2=y-22

b=2

Hence, the centre is​​ -4, 2

ii-​​ 

On comparing​​ x+42+y-22=20​​ with​​ x-a2+y-b2=r2, the radius of the circle comes out to be:

r2=20

r=20

r=25

​​ b- Solving equation of circle and equation of tangent simultaneously to get the coordinates of of point P.​​ 

x+2y+10=0

x=-2y-10

Substituting it in equation of circle.​​ 

x2+8x+y2-4y=0

-2y-102+8-2y-10+y2-4y=0

[4y2-2-2y10+-102]-16y-80+y2-4y=0

4y2+40y+100-16y-80+y2-4y=0

5y2+20y+20=0

y2+4y+4=0

y2+2y+2y+4=0

yy+2+2y+2=0

y+22=0

y=-2

To find the value of x, substitute the value of y in​​ x=-2y-10.

x=-2y-10=-2-2-10

x=-6

c-​​ 

The equation of tangent is​​ 

x+2y+10=0

2y= -x-10

y=-12 x-5 

So the gradient of tangent is​​ -12.​​ 

Since tangent’s and normal’s gradient are perpendicular to each other, so​​ 

mT×mN=-1

-12×mN=-1

Mn=2

Using point slope formula to get the equation of the normal.​​ 

(Since we have gradient of normal, which is​​ +2​​ and a point on normal, which is the centre of the circle-4, 2​​ , we may use the point-slope formula to find the equation of the normal.)

y-y1=mN(x-x1)

y-2=mN[x--4]

y-2=2x+4

y=2x+8+2

y=2x+10