Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2005 | January | Q#3

 

Question

A circle has equation

a.   By completing  the square, express the equation in the form

b.   Write down:

                           i.       the coordinates of the center of the circle;

                         ii.       the radius of the circle

c.   The line with equation  intersects the circle at the points P and Q.

                           i.       Show that the x-coordinates of P and Q satisfy the equation

                         ii.       Find the coordinates of P and Q.

Solution

a.
 

We have the algebraic formula;

For the given case we can rearrange the given equation and compare the given terms with the formula.

For terms containing  

For terms containing  

Therefore, we can deduce that;

To complete the square we can add and subtract the deduced value of ;

b.
 

If a circle is given by the equation  then the center of circle is at   and  is the radius of the circle.

Therefore, for the given case;


i.
 

Coordinates of center of circle .


ii.
 

Radius of circle 5.

c.    

Equation of the line given is;

Since line and the circle intersect, we find the coordinates of points of intersection.


i.
 

If equations of line and the circle are given;

ü From equation of line make  subject.

ü Substitute this into equation of the circle to get a quadratic equation in terms of  

ü To find the coordinates of points of intersection solve the quadratic equation for  and  substitute the found values of  into the equation of line to find  corresponding values of .

Equation of the line is;

Equation of the circle is;

Substituting value of  from equation of line into equation of circle;


ii.
 

We continue from (c:i) to find the coordinates of P and Q.

We solve the equation;

Now we have two options.

Two roots indicate that there are two intersection points.

We choose equation of the line;

For  

For  

Hence coordinates of  and .