Equation of a Circle

Edexcel IAL WMA12/01/P2 /June 2022/Q10 (Equations of Circles, Tangents, Discriminant)

The circle C has centre X(3, 5) and radius r​​ 

The line l has equation y = 2x + k , where k is a constant.​​ 

  • Show that l and C intersect when​​ 

5x2 + (4k - 26)x + k2 - 10k + 34 - r2 = 0

(3)

Given that l is a tangent to C,​​ 

  • show that 5r2 = (k + p) 2 , where p is a constant to be found.​​ 

(3)

Description: A diagram of a circle with a circle and a circle with a circle and a circle with a circle with a circle and a circle with a circle with a circle and a circle with a circle with

Description automatically generated

The line l​​ 

  • cuts the y-axis at the point A​​ 

  • touches the circle C at the point B​​ 

as shown in Figure 2.​​ 

Given that AB = 2r​​ 

  • find the value of k​​ 

(6)

SOLUTION​​ 

a-​​ 

We will solve the both equations (equation of circle and line) simultaneously to show that the given expression when both circle and line intersects.​​ 

Since​​ the circle C has centre X(3, 5) and radius r, the equation of circle will be​​ 

x-a2+y-b2=r2

x-32+y-52=r2

Now, substituting the value of y from the equation of line into the equation of a circle (y=2x + k).

x-32+2x+k-52=r2

x-32+(2x+k-5)2=r2

x2-6x+9+[4x2+22xk-5+k-52]=r2

5x2-6x+9+4kx-20x+k2-10k+25=r2

5x2+4kx-26x+k2-10k+34-r2=0

b-​​ 

Since the line l is tangent to C therefore the above equation, we found would have only one solution meaning thereby its discriminant would be equals 0.​​ 

So the equation of their intersection found in earlier part is

5x2+4kx-26x+k2-10k+34-r2=0

Where,​​ a=5; b=4k-26,  & c=k2-10k+34-r2 .

b2-4ac=0 

4k-262-45k2-10k+34-r2=0

4k2-24k26+262-20k2-10k+34-r2=0

16k2-209k+676-20k2+200k-680+20x2=0

-4k2-8k-4+20r2=0

-k2-2k-1+5r2=0

5r2=k2+2k+1

5r2=k+12  

c-​​ 

ABC forms a right angle triangle, applying the pythagoras theorem.

H2=P2+B2

AX2=2r2+r2

AX2=4r2+r2

AX2=5r2         (1)

On the otherhand, if we find the distance formula to find the distance between point A​​ (0, k)​​ and X​​ (3, 5), then we get​​ 

AX=x2-x12+y2-y12

AX2=3-02+5-k2

AX2=9+5-k2      (2)

On equating both equations (1) and (2),

5r2=9+5-k2    (3) 

Whereas we derived an expressiion in part (b) as​​ 

5r2=k+12   (4)

Now, equating equation (3) and (4), we get​​ 

k+12=9+5-k2

k2+2k+1=9+25-10k+k2

2k+10k=34-1

12k=33

k=3312

k=113