Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2020  FebMar  (P19709/12)  Q#12
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Question
A diameter of a circle C_{1} has endpoints at (−3, −5) and (7, 3).
a) Find an equation of the circle C_{1}.
The circle C_{1 }is translated by to give circle C_{2}, as shown in the diagram.
b) Find an equation of the circle C_{2}.
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 xcoordinates of R and S satisfy the equation
5x^{2} − 60x + 159 = 0
Solution
a)
We are given that points at (−3, −5) and (7, 3) makes endpoints of the diameter of a circle C_{1}.
If coordinates of end points of diameter, and are given;
ü Find the coordinates of midpoint 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 endpoints A(−3, −5) and B(7, 3) of diameter of the circle C_{1}.
To find the midpoint of a line we must have the coordinates of the endpoints of the line.
Expressions for coordinates of midpoint of a line joining points and ;
xcoordinate of midpoint of the line
ycoordinate of midpoint of the line
Hence;
xcoordinate of midpoint of the line
ycoordinate of midpoint 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 C_{1} is translated by to give circle C_{2 }and we are rquired to find the equation of the circle C_{2}.
Moving a curve without altering its shape is called a translation.
It is evident that given translation affects the center of the circle C_{1}; but not the radius .
Hence, to find the equation of the circle C_{2 }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 xdirection and units in the positive ydirection.
Original 
Transformed 
Translation Vector 
Movement 

Function 



units in positive xdirection units in positive ydirection 
Coordinates 


Therefore, the coordinates of new center of the circle C_{2} can be found as;
xcoordinate of
ycoordinate of
Hence, coordinates of the center of the circle C_{2} are and radius is .
Now, we can write equation of the circle C_{2 }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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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 xcoordinate or ycoordinate 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 xcoordinates of points R and S.
We have found the equation of the line joining R and S points by equating equations of the circles C_{1 }and C_{2}.
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;
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