Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2011 | January | Q#6

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Question

A circle has centre C(-3,1) and radius .

a.    

                    i.       Express the equation of the circle in the form

                  ii.       Hence find the equation of the circle in the form

where m, n and p are integers.

b.   The circle cuts the y-axis at the points A and B. Find the distance AB.

c.    

                    i.       Verify that the point D(-5,-2) lies on the circle.

                  ii.       Find the gradient of CD.

                 iii.       Hence find an equation of the tangent to the circle at the point D.

Solution

a.
 


i.
 

Expression for a circle with center at  and radius  is;

We are given that circle has centre C(-3,1) and radius .

Therefore, we can write equation of circle as;


ii.
 

We have found equation of the circle in (a:i) as;

b.
 

                            i.
 

We are required to find distance AB.

Expression to find distance between two given points  and is:

We need coordinates of points A and B to find the distance AB.

We are given that circle cuts the y-axis at points A and B. Therefore, points A and B are y-intercepts  of the circle.

The point  at which curve (or line) intercepts y-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

Therefore, we substitute  in equation of the circle obtained in (a:ii).

Now we have two options.

Two values of y indicate that there are two y-intercepts of the circle.

Therefore coordinates of A(0,-1) and B(0,3).

Now we can find the distance AB.

c.    

                            i.
 

If a given point lies on the circle then coordinates of point must satisfy equation of the circle.

We are given that point D has coordinates (-5,-2).

Therefore, we substitute coordinates (-5,-2) of point D in equation of the circle found in (a:ii);

Hence, point D (-5,-2) lies on the circle.


ii.
 

We are required to find gradient of CD.

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

We are given coordinates of both C(-3,1) and D(-5,-2). Therefore;


iii.
 

We are required to find equation of tangent to circle at point D.

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 already have coordinates of a point on tangent D(-5,-2). Therefore, we need slope of the tangent to write its equation.

We have shown that point D lies on the circle, therefore, CD is radius of the circle.

Radius is always perpendicular to tangent.

Therefore, tangent to circle at point D is perpendicular to CD. 

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

For the given case;

We have found in (b:ii) that;

Therefore;

Now we can write equation of the tangent.

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

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