Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2012 | January | Q#7

 

Question

A circle with centre C has .

a.   Express the equation in the form

b.   Write down:

                           i.       the coordinates of C;

                         ii.       the radius of the circle

c.   Sketch the circle.

d.   A line has the equation , where  is a constant.

                           i.       Show that the x-coordinates of any points of intersection of the line and the circle satisfy  the equation
.

                         ii.       The equation  has equal roots.  Show that;

                       iii.       Hence find the values of k for which the line is a tangent to the circle.

Solution

a.
 

We are given equation of the circle with center C as;

Expression for a circle with center at  and radius  is;

We can write the given equation of the circle in standard for as follows.

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.
 


i.
 

Expression for a circle with center at  and radius  is;

Comparing the equation from (a) with expression for circle;


i.       
Coordinates of center of circle .


ii.       
Radius of circle 5.

 

c.
 

d.    


i.
 

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 line is;

Equation of the curve is;

Substituting values of  in equation of the circle;


ii.
 

We are given the equation;

We are also given that this equation has equal roots.

For a quadratic equation , the expression for solution is;

Where  is called discriminant.

If , the equation will have two distinct roots.

If , the equation will have two identical/repeated roots.

If , the equation will have no roots.

The given equation is a quadratic equation and has equal roots (identical/repeated), therefore;


iii.
 

If the line is tangent to the circle then solution of equation which satisfies the x-coordinates of any  points of intersection of the line and the circle must yield a single solution/equal roots. 

We have seen from (ii) that it results in following equation.

To find the possible values of  we need to solve .

Therefore, we need to find the critical values for the equation;

Now we have two options.

Therefore, for these values of k equation will yield equal roots and hence line will intersect the circle  at a single point and, hence, will be tangent to circle.

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