Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2016 | June | Q#5



A circle with center C(5,-3) passes through the point A(-2,1).

a.   Find the equation of the circle in the form

b.   Given that AB is a diameter of the circle, find the coordinates of the point B.

c.   Find an equation of the tangent to the circle at the point A, giving your answer in the form  , where p, q and r are integers.

d.   The point T lies on the tangent to the circle at A such that AT=4. Find the length of CT.



We are given that circle has centre C(5,-3) and passes through the point A(-2,1).

Expression for a circle with center at  and radius  is;

We have coordinates of the center of the circle but we need to find radius.

It is evident that radius of the circle is AC with C(5,-3) and A(-2,1).

Expression to find distance between two given points  and is:

Hence, radius of circle is;

Therefore, equation of the circle is;


We are given that circle has centre C(5,-3) and passes through the point A(-2,1).

It is also given that point B(x,y) is such that AB is diameter, therefore, point B also lies on the circle.

We are required to find the coordinates of point B.

We translate the given information into a diagram as shown below.

It is evident that both vertical and horizontal distance from A to C must be equal to both vertical and  horizontal distance from C to B.


Since both vertical and horizontal distance from A to C must be equal to both vertical and horizontal  distance from C to B, we can extend from C to B by extending both distances to  corresponding coordinates of C.

For x-coordinate of point B, we need to move 7 units along positive x-axis away from point C,  therefore, horizontal distance is added to x-coordinate of point C.

Similarly, for y-coordinate of point B, we need to move 4 units along negative y-axis away from point  C, therefore, vertical distance is subtracted to y-coordinate of point C. 

Hence, coordinates of point B(12,-7).



We are required to find equation of the tangent to the circle at point A(-2,1). 

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).

We are given coordinates of a point on the tangent A(-2,1).

Therefore, we need slope of tangent. 

If two lines are perpendicular (normal) to each other, then product of their slopes  and  is; 

Radius is always perpendicular to tangent.


Hence, first we need to find slope of radius of the circle AC.

Expression for slope (gradient) of a line joining points  and ;

We are given that C(5,-3) and A(-2,1).


Now, with coordinates of point on the tangent A(-2,1) and slope of tangent , we can write  equation of the tangent.

Point-Slope form of the equation of the line is;


As we have seen in (c) that radius is always perpendicular to tangent, it is also evident that triangle  ATC is a right-angled triangle. Therefore, we can apply Pythagorean Theorem.

Pythagorean Theorem

For triangle QCD;

We have found that AC=radius= and we are given that , hence;