Past Papers’ Solutions  Assessment & Qualification Alliance (AQA)  AS & A level  Mathematics 6360  Pure Core 1 (6360MPC1)  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 xaxis.
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 xaxis 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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