2010 Paper 1 Q12 - equation of circles

Today is Deepavali. Best wishes to our Hindu friends!

We are now down to the last question of paper 1. So far, I am sure that most of you will agree that the paper is rather easy. Which is of course not good news too. This is because if many people found it easy, getting an A1 would not be so easy…

Anyway, let’s look at this last question

(i) Write down the equation of the circle with centre A(-3 , 2) and radius 5.

This circle intersects the y-axis at points P and Q.

(ii) Find the length of PQ.

A second circle, centre B, also passes through P and Q.

(iii) State the y-coordinate of B.

Given that the x-coordinate is positive and that the radius of the circle is √80, find

(iv) the x-coordinate of B

The equation of the circle, centre B which passes through P and Q may be written in the form

x² + y² +2gx + 2fy + c = 0.

(v)  State the value of g and f and find the value c.

Answer:

(i)  The first question is a real slam-dunk question. It really tests your memory of how the equation of a circle is.

So, let us recap. The equation of a circle, centre (h,k) and radius r, is given by

(x – h)² + (y – k)² = r²

Hence, the equation of the circle is  (x + 3)² + (y – 2)² = 5²

so,          x² + 6x + 9 + y² – 4y + 4 = 25

                x² + y² + 6x – 4y - 25 + 13 = 0

                x² + y² + 6x – 4y – 12 = 0

(ii) At the y-axis, x = 0. Therefore,   y² – 4y – 12 = 0

                                                                (y -  6)(y + 2) = 0

                                                                y = 6 or y = -2

Hence, P (0 , 6) and Q(0 , -2)  and the length PQ = 8 units.

(iii)  The y coordinate of B will be the midpoint of PQ and also the same y-coordinate of the first circle. Hence, the y-coordinate of B is 2.

(iv) For part (iv), we need to remember that the radius of the circle of centre B is the distance of B to any point in the circle. Since the radius of the second circle is √80, then, it means that PB = √80.

Let B be (p , q), then

PB² = p² + (6 – 2)² = 80

p² + 16 = 80

p² = 64

p = ±8 

But given that the x axis is positive, then the x-coordinate of B is 8.

(v)  Now that we have the coordinates of B(8, 2) and the radius of the circle, √80, we can put them into the equation of a circle as such:

(x – 8)² + (y – 2)² = 80

x² – 16x + 64 + y² – 4y + 4 – 80 = 0

x² + y² – 16x – 4y – 12 = 0

Hence, g = -8, f = -2 (which is really –h and –k respectively) and c = -12, which is really –(r² – h² – k²)