# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2020 | Feb-Mar | (P1-9709/12) | Q#12

Question

A diameter of a circle C1 has end-points at (3, 5) and (7, 3).

a)   Find an equation of the circle C1.

The circle C1 is translated by  to give circle C2, as shown in the diagram.

b)  Find an equation of the circle C2

The two circles intersect at points R and S.

c)   Show that the equation of the line RS is y = 2x + 13.

d)  Hence show that the x-coordinates of R and S satisfy the equation

5x2 60x + 159 = 0

Solution

a)

We are given that points at (3, 5) and (7, 3) makes end-points of the diameter of  a circle C1.

If coordinates of end points of diameter,  and  are given;

ü  Find the coordinates of mid-point of , this gives coordinates of center of the  circle

ü  Find the distance  or , this gives the radius of the circle

ü  Use the equation of the circle

Therefore, first we find the midpoint of given end-points A(3, 5) and B(7, 3) of  diameter of the circle C1.

To find the mid-point of a line we must have the coordinates of the end-points of the  line.

Expressions for coordinates of mid-point of a line joining points and ;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence, the point is the center of the circle .

Next, we find the radius of the circle by finding either distance CA or CB.

We choose to find CB.

Expression to find distance between two given points  and is:

Therefore, for CB;

We are given that B(7, 3) and we have found that . Hence;

Therefore, radius of the circle  is .

Now, using coordinates of center   and radius  of circle , we can  erite equation of the circle .

Expression for a circle with center at and radius  is;

Hence, for the circle ;

b)

We are given that the circle C1 is translated by  to give circle C2 and we are  rquired to find the equation of the circle C2.

Moving a curve without altering its shape is called a translation.

It is evident that given translation affects the center of the circle C1; but not  the radius  .

Hence, to find the equation of the circle C2 we need to find only coordinates of new  center of the circle.

We are given the translation as .

Translation of transforms the graph of into the graph of  .

Transformation of the function  into results from translation vector

Translation vector represents the move, units in the positive x-direction and units in the positive y-direction.

 Original Transformed Translation Vector Movement Function units in positive x-direction units in positive y-direction Coordinates

Therefore, the coordinates of new center of the circle C2 can be found as;

x-coordinate of

y-coordinate of

Hence, coordinates of the center of the circle C2 are  and radius is .

Now, we can write equation of the circle C2 as follows.

Expression for a circle with center at and radius  is;

c)

We are required to find the equation of the line RS where R and S are points of  intersection of the two circles.

To find the equation of the line either we need coordinates of the two points on the  line (Two-Point form of Equation of Line) or coordinates of one point on the line and  slope of the line (Point-Slope form of Equation of Line).

Therefore, we need to find coordinates of the points R and S.

If two lines (or a line and a curve) intersect each other at a point then that point lies  on both lines i.e. coordinates of that point have same values on both lines (or on  the line and the curve). Therefore, we can equate coordinates of both lines i.e.  equate equations of both the lines (or the line and the curve).

Equation of the circle  is;

Equation of the circle  is;

Equating both equations;

Since, we have not found x-coordinate or y-coordinate of the points of intersection  of circles  and but an equation; this equation represents the line joining these  points of intersection.

d)

We are required to find the equation satisfying x-coordinates of points R and S.

We have found the equation of the line joining R and S points by equating  equations of the circles C1 and C2.

Equation of the circle  is;

Equation of the circle  is;

We can substitute the equation found in (c) in any of the equation circles.

We choose;