Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2010 | January | Q#7

 

Question

A circle with centre C has .

a.   Find:

                           i.       the coordinates of C;

                         ii.       the radius of the circle

b.   Explain why the circle lies entirely below the x-axis.

c.   The point P with coordinates  lies outside the circle.

                           i.       Show that .

                         ii.       Hence show that

                       iii.       Find the possible values of k.

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 ;

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. 

b.
 

We have found in (a) that center of the circle is  and radius of the circle is .

It is evident that center of circle is 6 points below x-axis which is greater than radius of the circle which is 5.

c.
 

                            i.
 

We have found in (a:i) that  and we are also given that .

We are required to shoe that;

Expression to find distance between two given points  and is:

Therefore;


ii.
 

We are given that point P(5,k)  lies outside circle and we have found in (a:ii) that radius of the circle  is 5. Therefore;

 We have found from (b:i) that;

Therefore;


iii.
 

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, we have critical values as;

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If  (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
If
 (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

We recognize that given curve  , is a parabola opening upwards.

Therefore for the parabola   for;

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