Question

In the figure, the lines x=-8, y=q and the y-axis are tangent to the circle C. It is given that E(p,-2) is the centre of C. R is the point of intersection of the line x=-8 and y=q.
(a) (i) Find the value of p and q.
(ii) Find the value equation of circle C.
(b) G is the lowest point on C as shown in figure.
(i) Write down the coordinates of G.
(ii) P is a point on the circumference of C such that PG is perpendicular to RG. Find the coordinates of P

I don't know how to do b(ii)

Answer 1

For the figure, please refer to the attachment.

Answer 2

Okay i didnt work through the whole question, i assume u did. So i will just go straight to bII.

Knowing the coordinates of G, in b(i), you can find the equation of line RG, by finding the gradient using the 2 coordinates and consequently finding the y-intercept. After that you can get the equation of the perpendicular line, by manipulating the gradient.

You will thus get the equation of PG. Now you can solve for the values of P by using the equation of circle C and the equation of PG.

Answer 3

Okay.. Here it goes.

(ai)
p= -8/2 = -4
q= 4-2 = 2

(aii)
Equation of circle C is:
[x-(-4)]^2 + [y-(-2)]^2 = 4^2
(x+4)^2 + (y+2)^2 = 16

(bi)
G = (-4, -6)

(bii)
R = (-8, 2)
Slope of RG = [2-(-6)]/[(-8)-(-4)] = 8/-4 = -2
Slope of PG = -1/(-2) = 1/2
Equation of PG is:
(y+6)/(x+4) = 1/2
y = 1/2 x -4 -(1)

Put (1) into equation of circle C
(x+4)^2 + [(1/2 x -4)+2]^2 = 16
(x+4)^2 + (1/2 x -2)^2 = 16
5/4 x^2 + 6x + 20 = 16
5x^2 + 24x + 16 =0
x = -0.8 or x= -4 (rejected)
y= 1/2 (-0.8) -4 = -3.6

So, P = (-0.8, -3.6)

Is that correct?

Answer 4

looks good to me...
but hey, it's only me... i'm stupid...

Answer 5

You are not. But I am. Thanks!