Equation of a Circle

Edexcel IAL WMA12/01/P2 /Oct 2019/Q6 (Equation of Circles)

Figure 1 shows a sketch of a circle C with center N(4, - 1).

The line l with equation​​ y=12x​​ is a tangent to C at the point P.

Find

• The equation of line PN in the form​​ y=mx+c,​​ where m and c are constant,

(2)

• The equation of C.

(5)

SOLUTION​​ 

a- ​​ Using the equation of tangent​​ y=12x​​ to find the gradinet of line PN since it is normal to the line of tangent.

So comparing​​ y=12x​​ to the standard equation of a line which is​​ y=mx+c.

mN=12

Since​​ mPN⊥mT, the gradient of MN will be​​ 

mN×mT=-1

mMN×mT=-1

mT= -2

Now, using the point-slope formula. Substituting the gradient ​​ and point N(4, - 1) as the tangent passes through it.

y-y1=mx-x1

y-y1=mPN(x-x1)

y--1=-2x-4

y+1=-2x+8

y=-2x+7 

b- To find the equation of the circle, let us use the equation of a circle.​​ 

We need to substitute the values of centre of circle and radius.​​ 

For radius, let us find the distance PN, but for that we need point P whisch is the point of intersection of tangent and normal.​​ 

yT=12 x

yN= -2x+7

Equating both equations, to find the point of intersection, P.​​ 

12x=-2x+7

x= -4x+14

5x=14

x=145

Now, find the y-cordinate of point P.

yT=12 x

y=12 145

y=75

So,​​ P145,  75

Now, finding radius which is equal to PN.

r2=PN2=145-42+75+12

r2=PN2=365 

r2=365

At last, substitute the value of centre, N(4, - 1) and radius in the standard equation of circle.​​ 

x-a2+y-b2=r2

x-42+y+12=365