Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2015 | June | Q#4

 

Question

A circle with center C has the equation

a.   Express this equation in the form

b.    

                           i.       State the coordinates of C.

                         ii.       Find the radius of the circle, giving your answer in the form  .

c.    

                    i.       The point P with coordinates (4,k) lies on the circle. Find the possible values of k.

                  ii.       The points Q and R also lie on the circle, and the length of the chord QR is 2. Calculate the  shortest distance from C to the chord QR.

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;

We can write the equation of circle obtained in (a) in standard form as;

We can write equation in standard form as;

Comparing the equation from (a) with expression for circle we get coordinates of center of circle  .


ii.       
Radius of circle .

c.
 

It is evident that if point P(4,k) lies on the circle then distance of this point from the center of the  circle must be equal to radius of the circle.

Expression to find distance between two given points  and is:

Therefore for the given case;

We have found in (b:i) that coordinates of center of the circle are C(-1,3) and in (b:ii) that radius of  circle . Hence;

Now we have two options.

d.    

The perpendicular bisector of a chord in a circle passes through the center of the circle.

Therefore, it is evident from the diagram below that shortest possible distance from center of the circle to any chord of the circle is perpendicular distance from center of the circle to the chord.

It is also evident that triangle QCD is a right-angled triangle. Therefore, we can apply Pythagorean Theorem.

Pythagorean Theorem

For triangle QCD;

We have found that QC=radius= and , since CD is perpendicular bisector  of QR and we are given that QR=2.

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